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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s099556793
|
p02394
|
u105694406
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,584
| 135
|
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
w, h, x, y, r = map(int, input().split())
if x + r <= w and x - r <= 0 and y + r >= h and y - r <= 0:
print("Yes")
else:
print("No")
|
s190296408
|
Accepted
| 20
| 7,656
| 135
|
w, h, x, y, r = map(int, input().split())
if x + r <= w and x - r >= 0 and y + r <= h and y - r >= 0:
print("Yes")
else:
print("No")
|
s529147030
|
p04043
|
u189089176
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 201
|
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.
|
nums = list(map(int,input().split()))
for i in [5,7,5]:
if i in nums:
nums.remove(i)
else:
print("No")
exit()
if len(nums) == 0:
print("Yes")
else:
print("No")
|
s873738576
|
Accepted
| 17
| 2,940
| 329
|
nums = list(map(int,input().split()))
for i in [5,7,5]:
if i in nums:
nums.remove(i)
else:
print("NO")
exit()
if len(nums) == 0:
print("YES")
else:
print("NO")
|
s984826612
|
p02694
|
u966891144
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,188
| 199
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
def main():
x = int(input())
val = 100
year=1
while True:
val = val * 1.01
print(x,val)
if x <= val:
break
year += 1
print(year)
if __name__ == '__main__':
main()
|
s860746052
|
Accepted
| 22
| 9,164
| 187
|
def main():
x = int(input())
val = 100
year=1
while True:
val = int(val * 1.01)
if x <= val:
break
year += 1
print(year)
if __name__ == '__main__':
main()
|
s491590765
|
p03545
|
u374974389
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 606
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
num = list(input())
print(num)
ans = 7
for i in range(2 ** len(num)):
total = 0
ans_num = []
ans_op = []
for j in range(len(num)):
if((i >> j) & 1):
ans_num.append(num[j])
if(j >= 1):
ans_op.append('+')
total += int(num[j])
else:
ans_num.append(num[j])
if (j >= 1):
ans_op.append('-')
total -= int(num[j])
print(total)
print(ans_op)
if(total == ans):
print(ans_num[0]+ans_op[0]+ans_num[1]+ans_op[1]+ans_num[2]+ans_op[2]+ans_num[3]+'=7')
break
|
s090163371
|
Accepted
| 17
| 3,064
| 560
|
num = list(input())
ans = 7
for i in range(2 ** len(num)):
total = 0
ans_num = []
ans_op = []
for j in range(len(num)):
if((i >> j) & 1):
ans_num.append(num[j])
if(j >= 1):
ans_op.append('+')
total += int(num[j])
else:
ans_num.append(num[j])
if (j >= 1):
ans_op.append('-')
total -= int(num[j])
if(total == ans):
print(ans_num[0]+ans_op[0]+ans_num[1]+ans_op[1]+ans_num[2]+ans_op[2]+ans_num[3]+'=7')
break
|
s983081343
|
p03713
|
u474270503
| 2,000
| 262,144
|
Wrong Answer
| 206
| 4,192
| 291
|
There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle. Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece. Find the minimum possible value of S_{max} - S_{min}.
|
H, W=map(int, input().split())
m=float("inf")
for i in range(1,H-1):
S1=i*W
if (H-1)%2==0:
S2=W*((H-1)//2)
else:
S2=(H-i)*(W//2)
S3=H*W-S1-S2
if m>max(S1, S2, S3)-min(S1, S2, S3):
m=max(S1, S2, S3)-min(S1, S2, S3)
print(S1,S2,S3)
print(m)
|
s556362111
|
Accepted
| 467
| 3,064
| 613
|
H, W=map(int, input().split())
m=float("inf")
for i in range(1,H):
S1=i*W
S2=W*((H-i)//2)
S3=H*W-S1-S2
if m>max(S1, S2, S3)-min(S1, S2, S3):
m=max(S1, S2, S3)-min(S1, S2, S3)
S2=(H-i)*(W//2)
S3=H*W-S1-S2
if m>max(S1, S2, S3)-min(S1, S2, S3):
m=max(S1, S2, S3)-min(S1, S2, S3)
H,W=W,H
for i in range(1,H):
S1=i*W
S2=W*((H-i)//2)
S3=H*W-S1-S2
if m>max(S1, S2, S3)-min(S1, S2, S3):
m=max(S1, S2, S3)-min(S1, S2, S3)
S2=(H-i)*(W//2)
S3=H*W-S1-S2
if m>max(S1, S2, S3)-min(S1, S2, S3):
m=max(S1, S2, S3)-min(S1, S2, S3)
print(m)
|
s424159904
|
p02612
|
u950847221
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,104
| 121
|
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.
|
money=int(input('N'))
i=0
while i<=money:
if i*1000>=money:
break
else:
i+=1
print(i*1000-money)
|
s269824646
|
Accepted
| 30
| 9,024
| 118
|
money=int(input())
i=0
while i<=money:
if i*1000>=money:
break
else:
i+=1
print(i*1000-money)
|
s038679905
|
p03447
|
u761087127
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 70
|
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
X = int(input())
A = int(input())
B = int(input())
X =- A
print(X % B)
|
s300120700
|
Accepted
| 20
| 3,060
| 68
|
X = int(input())
A = int(input())
B = int(input())
X -= A
print(X%B)
|
s888363397
|
p03796
|
u612635771
| 2,000
| 262,144
|
Wrong Answer
| 153
| 9,848
| 70
|
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
import math
N = int(input())
print((math.factorial(N)) % 10 ** 9 + 7)
|
s264266276
|
Accepted
| 152
| 9,896
| 72
|
import math
N = int(input())
print((math.factorial(N)) % (10 ** 9 + 7))
|
s834170652
|
p03417
|
u035226531
| 2,000
| 262,144
|
Wrong Answer
| 2,135
| 529,840
| 793
|
There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region. The front and back sides of these cards can be distinguished, and initially every card faces up. We will perform the following operation once for each square contains a card: * For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square. It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed. Find the number of cards that face down after all the operations.
|
N, M = (int(i) for i in input().split())
tiles = [[True for i in range(M + 2)] for j in range(N + 2)]
print(tiles)
for i in range(1, N + 1):
for j in range(1, M + 1):
tiles[i - 1][j - 1] = not (tiles[i - 1][j - 1])
tiles[i - 1][j] = not (tiles[i - 1][j])
tiles[i - 1][j + 1] = not (tiles[i - 1][j + 1])
tiles[i][j - 1] = not (tiles[i][j - 1])
tiles[i][j] = not (tiles[i][j])
tiles[i][j + 1] = not (tiles[i][j + 1])
tiles[i + 1][j - 1] = not (tiles[i + 1][j - 1])
tiles[i + 1][j] = not (tiles[i + 1][j])
tiles[i + 1][j + 1] = not (tiles[i + 1][j + 1])
count = 0
for i in range(1, N + 1):
for j in range(1, M + 1):
print(tiles[i][j])
if not tiles[i][j]:
count = count + 1
print(count)
|
s433355619
|
Accepted
| 17
| 2,940
| 241
|
N, M = (int(i) for i in input().split())
if N == 1 and M == 1:
print(1)
elif N == 2 or M == 2:
print(0)
elif N == 1 or M == 1:
if N == 1:
print(M - 2)
else:
print(N - 2)
else:
print((N - 2) * (M - 2))
|
s332436739
|
p02608
|
u634880755
| 2,000
| 1,048,576
|
Wrong Answer
| 935
| 11,440
| 548
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
import math
N = int(input())
def xyzTriplets(N):
store = {}
upTo = math.sqrt(N)
for x in range(1, int(upTo) + 1):
for y in range(1, int(upTo) + 1):
for z in range(1, int(upTo) + 1):
res = x ** 2 + y ** 2 + z ** 2 + x * y + y * z + z * x
if (res not in store):
store[res] = 1
else:
store[res] += 1
for i in range(N):
if (i in store):
print(store[i])
else:
print(0)
xyzTriplets(N)
|
s706176073
|
Accepted
| 846
| 11,612
| 553
|
import math
N = int(input())
def xyzTriplets(N):
store = {}
upTo = math.sqrt(N)
for x in range(1, int(upTo) + 1):
for y in range(1, int(upTo) + 1):
for z in range(1, int(upTo) + 1):
res = x ** 2 + y ** 2 + z ** 2 + x * y + y * z + z * x
if (res not in store):
store[res] = 1
else:
store[res] += 1
for i in range(1, N+1):
if (i in store):
print(store[i])
else:
print(0)
xyzTriplets(N)
|
s382850609
|
p03352
|
u625554679
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 190
|
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
X = int(input())
max_num = 1
for b in range(2, 32):
num = 1
for p in range(1, 11):
num *= b
if num <= X and num >= max_num:
max_num = num
print(max_num)
|
s140734666
|
Accepted
| 17
| 2,940
| 190
|
X = int(input())
max_num = 1
for b in range(2, 32):
num = b
for p in range(2, 11):
num *= b
if num <= X and num >= max_num:
max_num = num
print(max_num)
|
s714898766
|
p03645
|
u431981421
| 2,000
| 262,144
|
Wrong Answer
| 689
| 51,744
| 369
|
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
import sys
N, M = map(int, input().split())
li = [list(map(int, input().split())) for n in range(M)]
a = []
b = []
for i in li:
if i[1] == N:
b.append(i[0])
if i[0] == N:
a.append(i[1])
if len(a) == 0 or len(b) == 0:
print("IMPOSSIBLE")
sys.exit()
for i in a:
if i in b:
print("IMPOSSIBLE")
sys.exit()
print("POSSIBLE")
|
s966011125
|
Accepted
| 752
| 67,472
| 314
|
import sys
from collections import defaultdict
d = defaultdict(int)
N, M = map(int, input().split())
li = [list(map(int, input().split())) for n in range(M)]
for i in li:
if i[1] == N:
d[i[0]] += 1
for i in li:
if i[0] == 1 and d[i[1]] != 0:
print("POSSIBLE")
sys.exit()
print("IMPOSSIBLE")
|
s354414878
|
p03385
|
u844196583
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,072
| 268
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
S = input()
a_list = list(S)
a_ct = 0
b_ct = 0
c_ct = 0
for i in range(0,3):
if a_list[i] == 'a':
a_ct = a_ct + 1
for i in range(0,3):
if a_list[i] == 'b':
b_ct = b_ct + 1
for i in range(0,3):
if a_list[i] == 'c':
c_ct = c_ct + 1
|
s045370844
|
Accepted
| 25
| 9,084
| 350
|
S = input()
a_list = list(S)
a_ct = 0
b_ct = 0
c_ct = 0
for i in range(0,3):
if a_list[i] == 'a':
a_ct = a_ct + 1
for i in range(0,3):
if a_list[i] == 'b':
b_ct = b_ct + 1
for i in range(0,3):
if a_list[i] == 'c':
c_ct = c_ct + 1
if a_ct == 1 and b_ct == 1 and c_ct == 1:
print('Yes')
else:
print('No')
|
s315660959
|
p04029
|
u725185906
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 44
|
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())
m=n-(n-1)
print((n/2)*(n+m))
|
s771945101
|
Accepted
| 17
| 2,940
| 49
|
n=int(input())
m=n-(n-1)
print(int((n/2)*(n+m)))
|
s729866036
|
p03644
|
u331997680
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 120
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N = int(input())
C = [2, 4, 8, 16, 32, 64]
for i in range(len(C)):
if C[i] < N:
continue
else:
print(C[i-1])
|
s052168312
|
Accepted
| 17
| 3,060
| 225
|
N = int(input())
C = [1, 2, 4, 8, 16, 32, 64]
for i in range(len(C)):
if N == C[i]:
print(N)
break
elif C[i] < N and N < 64:
continue
elif N >= 64:
print(64)
break
else:
print(C[i-1])
break
|
s114252757
|
p03997
|
u556589653
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 69
|
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)
|
s373120402
|
Accepted
| 19
| 3,060
| 72
|
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
|
s393593568
|
p02664
|
u194228880
| 2,000
| 1,048,576
|
Wrong Answer
| 91
| 9,400
| 269
|
For a string S consisting of the uppercase English letters `P` and `D`, let the _doctoral and postdoctoral quotient_ of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3. We have a string T consisting of `P`, `D`, and `?`. Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
|
t = input('')
a = ''
n = ''
for i in range(len(t)):
if t[i] == 'P':
n = 'D'
a += t[i]
if t[i] == 'D':
n = 'D'
a += t[i]
if t[i] == '?':
a+=n
print(a)
# DDDDDD -> 6
# PDPDPD -> 6
# PDPDPDP -> 3 + 3
# PDDDDDP -> 1 + 5
|
s144319831
|
Accepted
| 128
| 10,860
| 750
|
t = input('')
tlist = list(t)
for i in range(len(tlist)):
if tlist[i] == '?':
if i==0:
tlist[i] = 'D'
elif i==len(tlist)-1:
tlist[i] = 'D'
else:
if tlist[i-1] == 'P':
if tlist[i+1] == 'P':
tlist[i] = 'D'
elif tlist[i+1] == 'D':
tlist[i] = 'D'
elif tlist[i+1] == '?':
tlist[i] = 'D'
else:
if tlist[i+1] == 'P':
tlist[i] = 'D'
elif tlist[i+1] == 'D':
tlist[i] = 'P'
elif tlist[i+1] == '?':
tlist[i] = 'P'
print("".join(tlist))
|
s238890575
|
p02406
|
u518939641
| 1,000
| 131,072
|
Wrong Answer
| 40
| 7,564
| 108
|
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
n=input()
s=''
for i in range(int(n)+1):
s += (str(i) + ' ') if i%3==0 or '3' in str(i) else ''
print(s)
|
s463352615
|
Accepted
| 20
| 7,692
| 108
|
n=input()
s=''
for i in range(1,int(n)+1):
s += (' '+str(i)) if i%3==0 or '3' in str(i) else ''
print(s)
|
s394368306
|
p03623
|
u597017430
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
A = list(map(int, input().split()))
print(min(abs(A[0]-A[1]), abs(A[0]-A[2])))
|
s368201122
|
Accepted
| 17
| 2,940
| 91
|
A = list(map(int, input().split()))
print('A' if abs(A[0]-A[1]) <= abs(A[0]-A[2]) else 'B')
|
s294286329
|
p02865
|
u357230322
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 65
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n=int(input())
if n%2==0:
print((n/2)-1)
else:
print((n-1)/2)
|
s306644077
|
Accepted
| 17
| 2,940
| 76
|
n=int(input())
if n%2==0:
print(int((n/2)-1))
else:
print(int((n-1)/2))
|
s502695531
|
p03469
|
u243699903
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 31
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s = input()
print('2018'+s[5:])
|
s617900369
|
Accepted
| 17
| 2,940
| 31
|
s = input()
print('2018'+s[4:])
|
s510119157
|
p03369
|
u266171694
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 44
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
s = list(input())
print(700 + s.count('o'))
|
s689590772
|
Accepted
| 17
| 2,940
| 50
|
s = list(input())
print(700 + 100 * s.count('o'))
|
s175176978
|
p03251
|
u995102075
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 328
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N, M, X, Y = map(int, input().split())
x = sorted([int(i) for i in input().split()])
y = sorted([int(i) for i in input().split()])
Z_picklist1 = [range(X + 1, Y + 1)]
Z_picklist2 = [range(x[-1] + 1, y[0] + 1)]
if len(Z_picklist1 + Z_picklist2) != len(set(Z_picklist1 + Z_picklist2)):
print("No War")
else:
print("War")
|
s968467981
|
Accepted
| 17
| 3,060
| 276
|
N, M, X, Y = map(int, input().split())
x = sorted([int(i) for i in input().split()])
y = sorted([int(i) for i in input().split()])
flag = True
for i in range(X + 1, Y + 1):
if x[-1] < i and i <= y[0]:
flag = False
break
print("War" if flag else "No War")
|
s100457851
|
p03644
|
u474423089
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 149
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
def main():
n=int(input())
two_pow = 2
while two_pow < n:
two_pow *=2
print(two_pow/2)
if __name__ == '__main__':
main()
|
s395374969
|
Accepted
| 17
| 2,940
| 151
|
def main():
n=int(input())
two_pow = 2
while two_pow <= n:
two_pow *=2
print(two_pow//2)
if __name__ == '__main__':
main()
|
s030303870
|
p02612
|
u995163736
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,148
| 26
|
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)
|
s347459551
|
Accepted
| 27
| 9,012
| 44
|
n = int(input())
print((1000 - n%1000)%1000)
|
s843614614
|
p02412
|
u525685480
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,596
| 299
|
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
# -*- coding: utf-8 -*-
while True:
n,x = [int(i) for i in input().split(' ')]
count=0
if n+x ==0:
break
for i in range(n):
for j in range(n):
for k in range(n):
if i + j + k == x:
count +=1
print(count)
count=0
|
s652205722
|
Accepted
| 530
| 7,724
| 299
|
# -*- coding: utf-8 -*-
while True:
count=0
n,x = [int(a)for a in input().split(' ')]
if n+x==0:
break
for i in range(1,n-1):
for j in range(i+1,n):
for k in range(j+1,n+1):
if i + j + k ==x:
count += 1
print(count)
|
s561965477
|
p03993
|
u405256066
| 2,000
| 262,144
|
Wrong Answer
| 64
| 13,880
| 190
|
There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculate the number of the friendly pairs.
|
from sys import stdin
N = int(stdin.readline().rstrip())
A = [int(x) for x in stdin.readline().rstrip().split()]
ans = 0
for i in range(N):
if A[A[i]-1] == i:
ans += 1
print(ans)
|
s330578112
|
Accepted
| 71
| 13,880
| 195
|
from sys import stdin
N = int(stdin.readline().rstrip())
A = [int(x) for x in stdin.readline().rstrip().split()]
ans = 0
for i in range(N):
if A[A[i]-1] == i+1:
ans += 1
print(ans//2)
|
s244337760
|
p04044
|
u653005308
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 85
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
n,l=map(int,input().split())
list=[input() for i in range(n)]
list.sort()
print(list)
|
s758402229
|
Accepted
| 17
| 3,060
| 124
|
n,l=map(int,input().split())
list=[input() for i in range(n)]
list.sort()
ans=""
for word in list:
ans+=word
print(ans)
|
s612216092
|
p03698
|
u994521204
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
w=input()
if list(w)==list(set(w)):
print('yes')
else:print('no')
|
s088456770
|
Accepted
| 17
| 2,940
| 65
|
w=input()
if len(w)==len(set(w)):
print('yes')
else:print('no')
|
s189389043
|
p03679
|
u500297289
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 153
|
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.
|
""" AtCoder """
X, A, B = map(int, input().split())
if A >= B:
print("delicious")
elif (A + X) < B:
print("safe")
else:
print("dangerous")
|
s428356803
|
Accepted
| 18
| 2,940
| 154
|
""" AtCoder """
X, A, B = map(int, input().split())
if A >= B:
print("delicious")
elif (A + X) >= B:
print("safe")
else:
print("dangerous")
|
s894453163
|
p02390
|
u648117624
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,588
| 112
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
#coding: UTF-8
S = int(input('>'))
h = S//3600
m = (S%3600)//60
s = m%60
print(str(h)+':'+str(m)+':'+str(s))
|
s494786763
|
Accepted
| 20
| 5,580
| 61
|
x = int(input())
print(x//3600, x//60 % 60, x%60, sep=":")
|
s259634186
|
p02694
|
u068142202
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,164
| 135
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
x = int(input())
saving = 100
count = 0
while x >= saving:
saving += math.floor(saving * 0.01)
count += 1
print(count)
|
s470495028
|
Accepted
| 23
| 9,108
| 147
|
import math
x = int(input())
saving = 100
count = 0
while not x <= saving:
saving = math.floor(saving + saving * 0.01)
count += 1
print(count)
|
s651446980
|
p02697
|
u211706121
| 2,000
| 1,048,576
|
Wrong Answer
| 73
| 9,276
| 69
|
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(m):
print(i+1,2*m-i)
|
s610384672
|
Accepted
| 89
| 9,272
| 145
|
n,m=map(int,input().split())
k=n//2
c=0
while c<m:
d=c//2
if c%2==0:
print(k-d,k+1+d)
else:
print(d+1,n-1-d)
c+=1
|
s326608731
|
p02399
|
u641082901
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 89
|
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a, b = [int(x) for x in input().split()]
print("{0} {1} {2:.5f}".format(b//a, b%a, b/a))
|
s267057160
|
Accepted
| 20
| 5,612
| 89
|
a, b = [int(x) for x in input().split()]
print("{0} {1} {2:.5f}".format(a//b, a%b, a/b))
|
s686459288
|
p03469
|
u989345508
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 38
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s=list(input())
s[3]="8"
print(str(s))
|
s780002791
|
Accepted
| 17
| 2,940
| 42
|
s=list(input())
s[3]="8"
print("".join(s))
|
s432568003
|
p00002
|
u923573620
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,652
| 83
|
Write a program which computes the digit number of sum of two integers a and b.
|
import math
a, b = map(int, input().split())
print(str(int(math.log10(a+b)+1)))
|
s807722522
|
Accepted
| 20
| 5,588
| 107
|
while True:
try:
a, b = map(int, input().split(" "))
print(len(str(a + b)))
except:
break
|
s811519067
|
p03130
|
u346194435
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 3,064
| 459
|
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
def calc(x:list, y:int):
if x.count(y) == 0:
return -100
elif x.count(y) == 1:
return -1
elif x.count(y) == 2:
return 1
elif x.count(y) > 2:
return -100
loads = []
for _ in range(3):
loada, loadb = input().split()
loads.append(int(loada))
loads.append(int(loadb))
x = 0
for i in range(1,5):
print(i, calc(loads, i))
x = x + calc(loads, i)
if x == 0:
print('YES')
else:
print('NO')
|
s143746768
|
Accepted
| 17
| 3,064
| 430
|
def calc(x:list, y:int):
if x.count(y) == 0:
return -100
elif x.count(y) == 1:
return -1
elif x.count(y) == 2:
return 1
elif x.count(y) > 2:
return -100
loads = []
for _ in range(3):
loada, loadb = input().split()
loads.append(int(loada))
loads.append(int(loadb))
x = 0
for i in range(1,5):
x = x + calc(loads, i)
if x == 0:
print('YES')
else:
print('NO')
|
s057629978
|
p03730
|
u472534477
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 161
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
A,B,C = map(int,input().split())
flag = False
for i in range(1,B+1):
if (A*i) % B == C:
flag = True
if flag:
print("Yes")
else:
print("No")
|
s535750317
|
Accepted
| 17
| 3,064
| 161
|
A,B,C = map(int,input().split())
flag = False
for i in range(1,B+1):
if (A*i) % B == C:
flag = True
if flag:
print("YES")
else:
print("NO")
|
s623451460
|
p03478
|
u414050834
| 2,000
| 262,144
|
Wrong Answer
| 22
| 2,940
| 164
|
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())
c = 0
for i in range(1,n+1):
if a <= i//10000 + (i//1000)%10 + (i//100)%10 + (i//10)%10 + i&10 <= b:
c = c + i
print(c)
|
s773331104
|
Accepted
| 23
| 2,940
| 164
|
n,a,b = map(int,input().split())
c = 0
for i in range(1,n+1):
if a <= i//10000 + (i//1000)%10 + (i//100)%10 + (i//10)%10 + i%10 <= b:
c = c + i
print(c)
|
s968159069
|
p02280
|
u007270338
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,632
| 2,468
|
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.
|
#coding:utf-8
n = int(input())
T = [list(map(int, input().split())) for i in range(n)]
A = []
class binaryTree:
def __init__(self, x, y, z):
self.data = x
self.par = y
self.depth = z
def makeTree(self,tree):
parID = tree.data[0]
t = T[parID]
depth = tree.depth
if tree.par == -1:
tree.kind = "root"
elif t[1] == -1 and t[2] == -1:
tree.kind = "leaf"
else:
tree.kind = "international node"
if t[1] == -1 and t[2] == -1:
tree.deg = 0
tree.l = None
tree.r = None
if t[1] != -1:
tree.l = binaryTree(T[t[1]], parID, depth+1)
l_tree = tree.l
if t[2] != -1:
tree.deg = 2
l_tree.sib = t[2]
else:
tree.deg = 1
l_tree.sib = -1
self.makeTree(self, l_tree)
else:
tree.l = None
if t[2] != -1:
tree.r = binaryTree(T[t[2]], parID, depth+1)
r_tree = tree.r
if t[1] != -1:
tree.deg = 2
r_tree.sib = t[1]
else:
tree.deg = 1
r_tree.sib = -1
self.makeTree(self, r_tree)
else:
tree.r = None
def setHeight(tree):
h1 = h2 = 0
if tree.l != None:
h1 = setHeight(tree.l) + 1
if tree.r != None:
h2 = setHeight(tree.r) + 1
if tree.r == None and tree.l == None:
tree.hei = 0
return 0
tree.hei = max(h1,h2)
return max(h1, h2)
par = -1
depth = 0
tree = binaryTree(T[0], par,depth)
tree.sib = -1
binaryTree.makeTree(binaryTree, tree)
setHeight(tree)
ID = tree.data[0]
parent = tree.par
sibling = tree.sib
degree = tree.deg
depth = tree.depth
height = tree.hei
kind = tree.kind
def printer(tree):
ID = tree.data[0]
parent = tree.par
sibling = tree.sib
degree = tree.deg
depth = tree.depth
height = tree.hei
kind = tree.kind
print("node {}: parent = {}, sibling = {}, degree = {}, depth = {}, height = {}, {}"\
.format(ID, parent, sibling, degree, depth, height, kind))
if tree.l != None:
l_tree = tree.l
printer(l_tree)
if tree.r != None:
r_tree = tree.r
printer(r_tree)
printer(tree)
|
s642364611
|
Accepted
| 20
| 5,632
| 2,876
|
#coding:utf-8
n = int(input())
T = [list(map(int, input().split())) for i in range(n)]
A = [False for i in range(n)]
class binaryTree:
def __init__(self, node, x=-1):
self.node = node
self.p = x
def partialTree(self,tree,left, right):
node = tree.node
if left != -1:
if A[left]:
A[left].p = tree
tree.l = A[left]
else:
tree.l = binaryTree(left, node)
A[left] = tree.l
tree.l.p = tree
else:
tree.l = binaryTree(left, node)
tree.l.node = -1
if right != -1:
if A[right]:
A[right].p = tree
tree.r = A[right]
else:
tree.r = binaryTree(right, node)
A[right] = tree.r
tree.r.p = tree
else:
tree.r = binaryTree(right, node)
tree.r.node = -1
def searchParent():
p_tree = A[0]
while p_tree.p != -1:
p_tree = p_tree.p
return p_tree
def setHeight(tree):
h1 = h2 = 0
if tree.l.node != -1:
h1 = setHeight(tree.l) + 1
if tree.r.node != -1:
h2 = setHeight(tree.r) + 1
if tree.l.node == -1 and tree.r.node == -1:
tree.hei = 0
return 0
tree.hei = max(h1,h2)
return max(h1, h2)
def setDepth(tree,depth):
tree.dep = depth
depth += 1
if tree.l.node != -1:
setDepth(tree.l,depth)
if tree.r.node != -1:
setDepth(tree.r,depth)
def getSibling(tree):
if tree.p == -1:
return -1
if tree.p.l.node != -1 and tree.p.l.node != tree.node:
return tree.p.l.node
if tree.p.r.node != -1 and tree.p.r.node != tree.node:
return tree.p.r.node
return -1
def makeTree():
for i in range(n):
t = T[i]
node = t[0]
left = t[1]
right = t[2]
if A[node]:
tree = A[node]
else:
tree = binaryTree(node)
A[node] = tree
binaryTree.partialTree(binaryTree, tree,left,right)
makeTree()
p_tree = searchParent()
setHeight(p_tree)
depth = 0
setDepth(p_tree,depth)
for i in range(n):
tree = A[i]
tree.sib = getSibling(tree)
deg = 0
if tree.r.node != -1:
deg += 1
if tree.l.node != -1:
deg += 1
if tree.p == -1:
kind = "root"
else:
if deg == 0:
kind = "leaf"
else:
kind = "internal node"
ID = i
if tree.p == -1:
parent = -1
else:
parent = tree.p.node
sibling = tree.sib
degree = deg
depth = tree.dep
height = tree.hei
print("node {}: parent = {}, sibling = {}, degree = {}, depth = {}, height = {}, {}"\
.format(ID, parent, sibling, degree, depth, height, kind))
|
s331615857
|
p03997
|
u869282786
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 62
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a=int(input())
b=int(input())
c=int(input())
print((a+b)*c/2)
|
s672567229
|
Accepted
| 17
| 2,940
| 63
|
a=int(input())
b=int(input())
c=int(input())
print((a+b)*c//2)
|
s881576167
|
p03068
|
u727787724
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 133
|
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
n=int(input())
s=input()
k=int(input())
S=list(s)
for i in range(n):
if S[i]!=S[k-1]:
S[i]='*'
ans=','.join(S)
print(ans)
|
s930366567
|
Accepted
| 17
| 2,940
| 133
|
n=int(input())
s=input()
k=int(input())
S=list(s)
for i in range(n):
if S[i]!=S[k-1]:
S[i]='*'
ans=''.join(S)
print(ans)
|
s566290339
|
p03997
|
u241190800
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
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.
|
abh = [int(input()) for i in range(3)]
print((abh[0] + abh[1]) * abh[2] * 0.5)
|
s072632431
|
Accepted
| 17
| 2,940
| 149
|
data = [int(input()) for i in range(3)]
result = (data[0]+data[1])*data[2]*0.5
if result == int(result) :
print(int(result))
else :
print(result)
|
s711710561
|
p02397
|
u279483260
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,688
| 119
|
Write a program which reads two integers x and y, and prints them in ascending order.
|
def sort_num(x, y):
n = [x, y]
n.sort()
print(n[0], n[1])
x, y = map(int, input().split())
sort_num(x, y)
|
s318537189
|
Accepted
| 60
| 7,692
| 192
|
def sort_num(x, y):
n = [x, y]
n.sort()
print(n[0], n[1])
while True:
x, y = map(int, input().split())
if x == 0 and y == 0:
break
else:
sort_num(x, y)
|
s794483815
|
p02399
|
u780025254
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,692
| 80
|
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a, b = map(int, input().split())
d = a // b
r = a % b
f = a / b
print(d, r, f)
|
s195600901
|
Accepted
| 20
| 5,600
| 84
|
a, b = map(int, input().split())
print("{} {} {:.5f}".format(a // b, a % b, a / b))
|
s600569952
|
p02612
|
u984081384
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,096
| 69
|
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.
|
i = int(input())
if i % 1000:
print(0)
else:
print(1000-(i % 1000))
|
s183891871
|
Accepted
| 29
| 9,096
| 74
|
i = int(input())
if i % 1000 == 0:
print(0)
else:
print(1000-(i % 1000))
|
s474921674
|
p03545
|
u703442202
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 626
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
n = input()
s_num = len(n) -1
all_p_m = []
for i in range(2**s_num):
p_m_list = [False for i in range(s_num)]
for j in range(s_num):
if ((i >>j) & 1):
p_m_list[j] = True
all_p_m.append(p_m_list)
for p_m in all_p_m:
result = int(n[0])
for i in range(len(p_m)):
if p_m[i] == True:
result += int(n[i + 1])
else:
result -= int(n[i + 1])
if result == 7:
ans_p_m = p_m
ans = ["0" for i in range(7)]
for _ in range(7):
if _ %2 ==0:
ans[_] = n[_//2]
else:
if ans_p_m[_//2] == True:
ans[_] = "+"
else:
ans[_] = "-"
print("".join(ans))
|
s019004781
|
Accepted
| 18
| 3,064
| 642
|
n = input()
s_num = len(n) -1
all_p_m = []
for i in range(2**s_num):
p_m_list = [False for i in range(s_num)]
for j in range(s_num):
if ((i >>j) & 1):
p_m_list[j] = True
all_p_m.append(p_m_list)
for p_m in all_p_m:
result = int(n[0])
for i in range(len(p_m)):
if p_m[i] == True:
result += int(n[i + 1])
else:
result -= int(n[i + 1])
if result == 7:
ans_p_m = p_m
ans = ["0" for i in range(7)]
for _ in range(7):
if _ %2 ==0:
ans[_] = n[_//2]
else:
if ans_p_m[_//2] == True:
ans[_] = "+"
else:
ans[_] = "-"
ans.append("=7")
print("".join(ans))
|
s988377593
|
p02602
|
u312158169
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 31,704
| 339
|
M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term.
|
n,k = map(int,input().split())
a = [int(x) for x in input().split()]
num = n-k
ans = [0] * (num+1)
ans[0] = 1
for i in range(num+1):
ans[0] *= a[i]
for i in range(1,num+1):
ans[i] = ans[i-1]/ a[i-1] * a[i+k-1]
print(ans)
for i in range(1,num+1):
if ans[i-1] >= ans[i]:
print("No")
else:
print("Yes")
|
s518467422
|
Accepted
| 168
| 31,612
| 184
|
n,k = map(int,input().split())
a = [int(x) for x in input().split()]
num = n-k
for i in range(1,num+1):
if a[i-1] >= a[i+k-1]:
print("No")
else:
print("Yes")
|
s693010163
|
p03370
|
u536325690
| 2,000
| 262,144
|
Wrong Answer
| 32
| 3,060
| 224
|
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
n, x = map(int, input().split())
m = [int(input()) for i in range(n)]
rest = x - sum(m)
min_ = min(m)
print(m)
count = n
while True:
rest -= min_
if rest > 0:
count += 1
else:
break
print(count)
|
s197072277
|
Accepted
| 36
| 3,060
| 274
|
n, x = map(int, input().split())
m = [int(input()) for i in range(n)]
rest = x - sum(m)
min_ = min(m)
count = n
if rest < 0:
count = 0
else:
while True:
rest -= min_
if rest >= 0:
count += 1
else:
break
print(count)
|
s851847875
|
p02612
|
u430937688
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,144
| 52
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N - ((N - 1) // 1000) * 1000)
|
s846370383
|
Accepted
| 27
| 9,144
| 56
|
N = int(input())
print(((N - 1) // 1000 + 1) * 1000 - N)
|
s402107641
|
p03997
|
u246661425
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 69
|
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(h/2 * (a+b))
|
s099248428
|
Accepted
| 17
| 2,940
| 75
|
a = int(input())
b = int(input())
h = int(input())
print(int((a+b) * h /2))
|
s810760772
|
p02600
|
u645661835
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,156
| 115
|
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
X=int(input())
initial=400
kyu=8
for add in range(1,9):
if X<initial+add*200:
print(kyu)
break
kyu-=1
|
s795170459
|
Accepted
| 34
| 8,852
| 130
|
X=int(input())
initial=400
kyu=8
for add in range(1,9):
if X<initial+add*200:
print(int(kyu))
break
kyu-=1
|
s752368903
|
p00020
|
u184989919
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,428
| 76
|
Write a program which replace all the lower-case letters of a given text with the corresponding captital letters.
|
def Capitalize():
for i in sys.stdin:
print(i.upper())
|
s568271111
|
Accepted
| 30
| 7,276
| 22
|
print(input().upper())
|
s862310947
|
p03095
|
u571969099
| 2,000
| 1,048,576
|
Wrong Answer
| 544
| 3,188
| 153
|
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.
|
n=int(input())
s=input()
l=n
for i in range(n):
a=set()
for j,k in enumerate(s[i:]):
a.add(k)
if len(a)!=j+1:
break
l+=1
print(l)
|
s412713855
|
Accepted
| 35
| 3,188
| 143
|
n=int(input())
s=input()
a={}
for i in s:
if i in a:
a[i]+=1
else:
a[i]=1
j=1
for k in a:
i=a[k]
j*=i+1
print((j-1)%(10**9+7))
|
s461853109
|
p03860
|
u588633699
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 34
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
s = input().split()
print(s[2][0])
|
s706597876
|
Accepted
| 17
| 2,940
| 42
|
s = input().split()
print('A'+s[1][0]+'C')
|
s498833001
|
p03351
|
u008357982
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 91
|
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());print('YNeos'[abs(a-c)>d or abs(a-b)>d and abs(b-c)>d::2])
|
s624225632
|
Accepted
| 17
| 2,940
| 101
|
a,b,c,d=map(int,input().split())
print("Yes" if abs(c-a)<=d or abs(b-a)<=d and abs(c-b)<=d else "No")
|
s700159973
|
p03068
|
u706884679
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 192
|
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
N = int(input())
S = list(input())
K = int(input())
x = S[K-1]
s = [i for i, t in enumerate(S) if t == x]
for i in range(N):
if i in s:
S[i-1] = '*'
else:
pass
A = "".join(S)
print(A)
|
s341507859
|
Accepted
| 18
| 3,060
| 194
|
N = int(input())
S = list(input())
K = int(input())
x = S[K-1]
s = [i for i, t in enumerate(S) if t == x]
for i in range(N):
if i not in s:
S[i] = '*'
else:
pass
A = "".join(S)
print(A)
|
s776632021
|
p03910
|
u335540120
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,316
| 185
|
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.
|
N = int(input())
ans_max = 0
while ans_max * (ans_max + 1) / 2 < N:
ans_max += 1
num_remove = ans_max - N
for i in range(1, ans_max + 1):
if i != num_remove:
print(i)
|
s115787722
|
Accepted
| 23
| 3,316
| 206
|
N = int(input())
ans_max = 0
while ans_max * (ans_max + 1) / 2 < N:
ans_max += 1
num_remove = ans_max * (ans_max + 1) // 2 - N
for i in range(1, ans_max + 1):
if i != num_remove:
print(i)
|
s173977445
|
p02619
|
u729133443
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,060
| 1
|
Let's first write a program to calculate the score from a pair of input and output. You can know the total score by submitting your solution, or an official program to calculate a score is often provided for local evaluation as in this contest. Nevertheless, writing a score calculator by yourself is still useful to check your understanding of the problem specification. Moreover, the source code of the score calculator can often be reused for solving the problem or debugging your solution. So it is worthwhile to write a score calculator unless it is very complicated.
|
0
|
s765183144
|
Accepted
| 31
| 10,164
| 145
|
d,*s,=map(int,open(i:=0).read().split())
c,*l=[0]*27
for v in s[-d:]:
i+=1
l[v-1]=i
print(c:=c+s[i*26+v-1]-sum(x*(i-y)for x,y in zip(s,l)))
|
s523684365
|
p03069
|
u528470578
| 2,000
| 1,048,576
|
Wrong Answer
| 70
| 3,560
| 243
|
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
N = int(input())
S = str(input())
new_S = ""
for i in reversed(S):
new_S += i
for i in range(N):
if new_S[i] == ".":
S_new = S[i:]
break
else:
S_new = ""
print(min(S_new.count('
|
s662098088
|
Accepted
| 93
| 11,368
| 251
|
N = int(input())
S = str(input())
suu = []
s_count = S.count('.')
k_count = 0
suu.append(s_count + k_count)
for i in range(N):
if S[i] == "#":
k_count += 1
else:
s_count -= 1
suu.append(k_count + s_count)
print(min(suu))
|
s230065851
|
p02397
|
u656153606
| 1,000
| 131,072
|
Wrong Answer
| 50
| 7,648
| 138
|
Write a program which reads two integers x and y, and prints them in ascending order.
|
while True:
a, b = [int(i) for i in input().split()]
if a == 0 and b == 0:
break
print(a, b) if a > b else print(b, a)
|
s019433573
|
Accepted
| 40
| 8,248
| 247
|
list = []
while True:
a, b = [int(i) for i in input().split()]
if a == 0 and b == 0:
break
if a < b:
list.append([a,b])
else:
list.append([b,a])
for i in range(len(list)):
print(list[i][0], list[i][1])
|
s891794663
|
p03693
|
u075409829
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 94
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r, g, b = map(int, input().split())
if (g*10+b)%4 == 0:
print("Yes")
else:
print("No")
|
s620675957
|
Accepted
| 17
| 2,940
| 94
|
r, g, b = map(int, input().split())
if (g*10+b)%4 == 0:
print("YES")
else:
print("NO")
|
s601542867
|
p02600
|
u354862173
| 2,000
| 1,048,576
|
Wrong Answer
| 39
| 9,972
| 628
|
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
import sys
from string import ascii_letters as ascii
def Ss():
return sys.stdin.readline().rstrip()
def Is():
return map(int, Ss())
def get_kyu(score):
if score >= 400 and score <= 599:
return 8
if score >= 600 and score <= 799:
return 7
if score >= 800 and score <= 999:
return 6
if score >= 1000 and score <= 1199:
return 5
if score >= 1200 and score <= 1399:
return 4
if score >= 1400 and score <= 1599:
return 3
if score >= 1600 and score <= 1799:
return 2
if score >= 1800 and score <= 1999:
return 1
return None
score = int(Ss())
print(score)
print(get_kyu(score))
|
s960821252
|
Accepted
| 42
| 9,968
| 615
|
import sys
from string import ascii_letters as ascii
def Ss():
return sys.stdin.readline().rstrip()
def Is():
return map(int, Ss())
def get_kyu(score):
if score >= 400 and score <= 599:
return 8
if score >= 600 and score <= 799:
return 7
if score >= 800 and score <= 999:
return 6
if score >= 1000 and score <= 1199:
return 5
if score >= 1200 and score <= 1399:
return 4
if score >= 1400 and score <= 1599:
return 3
if score >= 1600 and score <= 1799:
return 2
if score >= 1800 and score <= 1999:
return 1
return None
score = int(Ss())
print(get_kyu(score))
|
s695313119
|
p04011
|
u664481257
| 2,000
| 262,144
|
Wrong Answer
| 81
| 7,544
| 689
|
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.
|
# -*- coding: utf-8 -*-
# !/usr/bin/env python
# vim: set fileencoding=utf-8 :
def read_input():
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
return N, K, X, Y
def check_sum(N, K, X, Y):
if N > K:
return K * X + (N-K)*Y
else:
return N * X
if __name__ == "__main__":
import doctest
doctest.testmod()
N, K, X, Y = read_input()
check_sum(N, K, X, Y)
|
s050355978
|
Accepted
| 22
| 3,064
| 705
|
# -*- coding: utf-8 -*-
# !/usr/bin/env python
# vim: set fileencoding=utf-8 :
def read_input():
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
return N, K, X, Y
def check_sum(N, K, X, Y):
if N > K:
return K * X + (N-K)*Y
else:
return N * X
if __name__ == "__main__":
#
# doctest.testmod()
N, K, X, Y = read_input()
print(check_sum(N, K, X, Y))
|
s674245676
|
p03005
|
u373047809
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 57
|
Takahashi is distributing N balls to K persons. If each person has to receive at least one ball, what is the maximum possible difference in the number of balls received between the person with the most balls and the person with the fewest balls?
|
n, k = map(int, input().split())
print([0, 0, n-k][k!=1])
|
s637409815
|
Accepted
| 18
| 2,940
| 37
|
print(eval(input().replace(' ','%')))
|
s679151941
|
p03696
|
u690037900
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 320
|
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
import sys
input=sys.stdin.readline
N = int(input())
S = list(input())
f = 0
b = 0
count = 0
for i in range(N - 1, -1, -1):
if S[i] == '(':
if count > 0:
count -= 1
else:
b += 1
else:
count += 1
print('(' * count, end='')
print(*S, sep='', end='')
print(')' * b)
|
s729340965
|
Accepted
| 17
| 3,064
| 286
|
N = int(input())
S = list(input())
f = 0
b = 0
count = 0
for i in range(N - 1, -1, -1):
if S[i] == '(':
if count > 0:
count -= 1
else:
b += 1
else:
count += 1
print('(' * count, end='')
print(*S, sep='', end='')
print(')' * b)
|
s108153327
|
p03369
|
u952022797
| 2,000
| 262,144
|
Wrong Answer
| 161
| 13,260
| 463
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
# -*- coding: utf-8 -*-
import sys
import copy
import collections
from bisect import bisect_left
from bisect import bisect_right
from collections import defaultdict
from heapq import heappop, heappush
import numpy as np
import statistics
from statistics import mean, median,variance,stdev
import math
def main():
s = input()
tmp = 0
for i in range(len(s)):
if i == "o":
tmp += 1
print(700 + (tmp * 100))
if __name__ == "__main__":
main()
|
s119050066
|
Accepted
| 161
| 13,260
| 466
|
# -*- coding: utf-8 -*-
import sys
import copy
import collections
from bisect import bisect_left
from bisect import bisect_right
from collections import defaultdict
from heapq import heappop, heappush
import numpy as np
import statistics
from statistics import mean, median,variance,stdev
import math
def main():
s = input()
tmp = 0
for i in range(len(s)):
if s[i] == "o":
tmp += 1
print(700 + (tmp * 100))
if __name__ == "__main__":
main()
|
s054985198
|
p03251
|
u953379577
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,136
| 273
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
n,m,a,b = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
x.sort()
y.sort()
if x[-1]<y[0]:
for i in range(x[-1]+1,y[0]+1):
if a<i<b:
print("No War")
else:
print("War")
else:
print("War")
|
s482585352
|
Accepted
| 29
| 9,140
| 291
|
n,m,a,b = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
x.sort()
y.sort()
if x[-1]<y[0]:
for i in range(x[-1]+1,y[0]+1):
if a<i<b:
print("No War")
break
else:
print("War")
else:
print("War")
|
s157396192
|
p03943
|
u244836567
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,116
| 115
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a,b,c=input().split()
a=int(a)
b=int(b)
c=int(c)
ls=[a,b,c]
ls.sort()
if a+b==c:
print("Yes")
else:
print("No")
|
s969322870
|
Accepted
| 28
| 9,116
| 127
|
a,b,c=input().split()
a=int(a)
b=int(b)
c=int(c)
ls=[a,b,c]
ls.sort()
if ls[0]+ls[1]==ls[2]:
print("Yes")
else:
print("No")
|
s813656753
|
p03796
|
u479638406
| 2,000
| 262,144
|
Wrong Answer
| 29
| 2,940
| 72
|
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
n = int(input())
v = 1
for _ in range(n):
v = v*v
print(v%(1e9+7))
|
s208019434
|
Accepted
| 72
| 2,940
| 109
|
n = int(input())
v = 1
for i in range(n):
v = v*(i+1)
if v >= 1e9+7:
v = v%(int(1e9)+7)
print(v)
|
s073283548
|
p03759
|
u235376569
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 104
|
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c = [int(x) for x in input().rstrip().split()]
if b-c == c-b:
print("YES")
else:
print("NO")
|
s325940368
|
Accepted
| 19
| 2,940
| 104
|
a,b,c = [int(x) for x in input().rstrip().split()]
if b-a == c-b:
print("YES")
else:
print("NO")
|
s959672181
|
p02833
|
u077291787
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 139
|
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
# ABC148E - Double Factorial
def main():
N = int(input())
if N % 2 == 1:
print(0)
if __name__ == "__main__":
main()
|
s257299697
|
Accepted
| 17
| 2,940
| 253
|
# ABC148E - Double Factorial
def main():
N = int(input())
if N % 2:
print(0)
else:
N //= 10
ans = N
while N:
N //= 5
ans += N
print(ans)
if __name__ == "__main__":
main()
|
s845161816
|
p03351
|
u755944418
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 130
|
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,x = map(int, input().split())
if abs(c-a) <= x or (abs(c-b) <= x and abs(b-a) <= x):
print('YES')
else:
print('NO')
|
s270525453
|
Accepted
| 17
| 2,940
| 130
|
a,b,c,x = map(int, input().split())
if abs(c-a) <= x or (abs(c-b) <= x and abs(b-a) <= x):
print('Yes')
else:
print('No')
|
s258519361
|
p02388
|
u451779396
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,312
| 14
|
Write a program which calculates the cube of a given integer x.
|
input("x")*3
|
s206780586
|
Accepted
| 20
| 7,672
| 27
|
x=int(input())
print(x**3)
|
s584002802
|
p03067
|
u452844010
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 3,060
| 180
|
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
import sys
str = input()
if str[0] < str[2]:
if str[1] > str[2]:
print('Yes')
exit(0)
if str[0] > str[2]:
if str[1] < str[2]:
print('Yes')
exit(0)
print('No')
|
s744694762
|
Accepted
| 17
| 3,064
| 252
|
import sys
str = input().split()
str[0] = int(str[0])
str[1] = int(str[1])
str[2] = int(str[2])
if str[0] < str[2]:
if str[1] > str[2]:
print('Yes')
exit(0)
if str[0] > str[2]:
if str[1] < str[2]:
print('Yes')
exit(0)
print('No')
|
s102396167
|
p02601
|
u330661451
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,205
| 9,252
| 252
|
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
def main():
a,b,c = map(int,input().split())
k = int(input())
i = 0
while a >= b:
a *= 2
i += 1
while b >= c:
b *= 2
i += 1
print("Yes" if i <= k else "No")
if __name__ == '__main__':
main()
|
s110683976
|
Accepted
| 32
| 9,096
| 252
|
def main():
a,b,c = map(int,input().split())
k = int(input())
i = 0
while a >= b:
b *= 2
i += 1
while b >= c:
c *= 2
i += 1
print("Yes" if i <= k else "No")
if __name__ == '__main__':
main()
|
s653926194
|
p03049
|
u760794812
| 2,000
| 1,048,576
|
Wrong Answer
| 39
| 3,700
| 453
|
Snuke has N strings. The i-th string is s_i. Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of `AB` in the resulting string.
|
N = int(input())
counter = 0
S = []
for s in range(N):
s = input()
S.append(s)
S0 = S[0]
a0 = S0.count('A')
b0 = S0.count('B')
counter += min(a0,b0)
a_counter = 0
if a0 > b0:
a_counter +=1
for i in range(N-1):
ST = S[i+1]
a = ST.count('A')
b = ST.count('B')
if (a_counter == 1) and (b > 1):
counter +=1
b = b -1
counter += min(a,b)
if a > b:
a_couter = 1
else:
a_counter = 0
print(counter)
|
s641023968
|
Accepted
| 46
| 9,192
| 348
|
n = int(input())
xa = 0
by = 0
ba = 0
counter = 0
for _ in range(n):
s = input()
counter += s.count('AB')
if s[0]=='B':
if s[-1]=='A':
ba += 1
else:
by += 1
elif s[-1] == 'A':
xa += 1
if ba == 0:
counter += min(xa,by)
else:
if xa+by== 0:
counter += ba -1
else:
counter += ba + min(xa,by)
print(counter)
|
s487606001
|
p03567
|
u403753892
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 89
|
Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the string `AC` as the verdict to a program.
|
import re
s = input()
m = re.match("AC",s)
if m:
print("Yes")
else:
print("No")
|
s213137794
|
Accepted
| 19
| 3,188
| 91
|
import re
s = input()
m = re.search("AC",s)
if m:
print("Yes")
else:
print("No")
|
s165725881
|
p03048
|
u373046572
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 2,940
| 189
|
Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?
|
r,g,b,n = map(int,input().split())
cnt = 0
for i in range(n // r + 1):
for j in range(n // g + 1):
c = n - i * r - j * g
if c % b == 0:
cnt += 1
print(cnt)
|
s018631133
|
Accepted
| 1,912
| 2,940
| 228
|
r,g,b,n = map(int,input().split(" "))
cnt = 0
for i in range(n // r + 1):
for j in range(n // g + 1):
c = n - i * r - j * g
if c < 0:
break
if c % b == 0:
cnt += 1
print(cnt)
|
s964557839
|
p02694
|
u470189643
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,156
| 134
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
X=int(input())
yen=100
for i in range(3760):
yen=math.floor(yen*1.01)
if yen>X:
break
print(i+1)
|
s673355159
|
Accepted
| 22
| 9,160
| 145
|
import math
X=int(input())
yen=int(100)
for i in range(3800):
yen=int(math.floor(yen*1.01))
if yen>=X:
break
print(i+1)
|
s184322541
|
p03679
|
u487288850
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,144
| 114
|
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 b<=a:
print('delicious')
elif b<=x:
print("safe")
else:
print("dangerous")
|
s088428273
|
Accepted
| 25
| 8,960
| 116
|
x,a,b=map(int,input().split())
if b<=a:
print('delicious')
elif b-a<=x:
print("safe")
else:
print("dangerous")
|
s516867516
|
p03720
|
u858670323
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 204
|
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
n,m = map(int,input().rstrip().split(' '))
G = [[]]*n
for i in range(m):
a,b = map(int,input().rstrip().split(' '))
a-=1
b-=1
G[a].append(b)
G[b].append(a)
for i in range(n):
print(len(G[i]))
|
s965643661
|
Accepted
| 17
| 3,060
| 221
|
n,m = map(int,input().rstrip().split(' '))
G = [[] for i in range(n)]
for i in range(m):
a,b = map(int,input().rstrip().split(' '))
a-=1
b-=1
G[a].append(b)
G[b].append(a)
for i in range(n):
print(len(G[i]))
|
s068113076
|
p03370
|
u912652535
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 121
|
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
n , x = map(int,input().split())
m = [int(input()) for i in range(n)]
print(m)
total = sum(m)
print((x-total)//min(m)+n)
|
s033526261
|
Accepted
| 18
| 2,940
| 113
|
n , x = map(int,input().split())
m = [int(input()) for i in range(n)]
total = sum(m)
print((x-total)//min(m)+n)
|
s301159521
|
p03456
|
u839953865
| 2,000
| 262,144
|
Wrong Answer
| 25
| 3,060
| 180
|
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())
k=int(a+b)
c=0
for i in range(1,k):
if k%(k-i)==0:
c=k-i
break
print(k)
print(c)
if k==c*c:
print("Yes")
else:
print("No")
|
s977067009
|
Accepted
| 29
| 2,940
| 127
|
a,b=map(str,input().split())
k=int(a+b)
c=0
for i in range(1,k):
if k==i*i:
print("Yes")
exit()
print("No")
|
s954932904
|
p03456
|
u198058633
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 97
|
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.
|
import math
sq=math.sqrt(int("".join(input().split())))
print("YES" if sq.is_integer() else "NO")
|
s217556097
|
Accepted
| 17
| 2,940
| 78
|
s=int(input().replace(" ", ""))
print("Yes" if int(s**0.5)**2 == s else "No")
|
s415538180
|
p02406
|
u427088273
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,660
| 82
|
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
num = int(input())
print(' '.join([str(i) for i in range(1,num+1) if i % 3 == 0]))
|
s639166253
|
Accepted
| 40
| 7,868
| 96
|
print(' ' + ' '.join(str(i) for i in range(1, int(input()) + 1) if i % 3 == 0 or '3' in str(i)))
|
s227772556
|
p03693
|
u430483125
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 106
|
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 = input().split()
number = int(r+g+b)
if number % 4 ==0:
print('Yes')
else:
print('No')
|
s009881950
|
Accepted
| 17
| 2,940
| 106
|
r, g, b = input().split()
number = int(r+g+b)
if number % 4 ==0:
print('YES')
else:
print('NO')
|
s472997377
|
p02796
|
u340781749
| 2,000
| 1,048,576
|
Wrong Answer
| 241
| 18,260
| 321
|
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 sys
from operator import itemgetter
n = int(input())
robots = []
for line in sys.stdin:
X, L = map(int, line.split())
l, r = X - L, X + L
robots.append((l, r))
robots.sort(key=itemgetter(1))
curr = -1
ans = 0
for l, r in robots:
if curr > l:
continue
ans += 1
curr = r
print(ans)
|
s824490819
|
Accepted
| 250
| 18,260
| 330
|
import sys
from operator import itemgetter
n = int(input())
robots = []
for line in sys.stdin:
X, L = map(int, line.split())
l, r = X - L, X + L
robots.append((l, r))
robots.sort(key=itemgetter(1))
curr = -(10 ** 10)
ans = 0
for l, r in robots:
if curr > l:
continue
ans += 1
curr = r
print(ans)
|
s579504405
|
p04043
|
u083593845
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 134
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a,b,c = map(int, input().split())
s = {a,b,c}
list1 = [a,b,c]
if s == {5,7} and sum(list1) == 17:
print("Yes")
else :
print("No")
|
s258957569
|
Accepted
| 19
| 3,060
| 133
|
a,b,c = map(int, input().split())
s = {a,b,c}
list1 = [a,b,c]
if s == {5,7} and sum(list1) == 17:
print("YES")
else :
print("NO")
|
s206784741
|
p02612
|
u556594202
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 8,876
| 31
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N//1000)
|
s072730618
|
Accepted
| 27
| 9,052
| 73
|
N = int(input())
if N%1000==0:
print(0)
else:
print(1000-N%1000)
|
s507607524
|
p03478
|
u843318925
| 2,000
| 262,144
|
Wrong Answer
| 37
| 3,060
| 172
|
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())
total = 0
for i in range(n):
s = str(i + 1)
x = sum(list(map(int, s)))
if x >= a & x <= b:
total += x
print(total)
|
s698322923
|
Accepted
| 35
| 3,060
| 161
|
n, a, b = map(int, input().split())
total = 0
for i in range(1, n + 1):
s = str(i)
if a <= sum(list(map(int, s))) <= b:
total += i
print(total)
|
s899955683
|
p03659
|
u875361824
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 136,876
| 309
|
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
def main():
N = int(input().strip())
a = list(map(int, input().strip().split()))
min_v = float("inf")
for i in range(1, N):
print(a[:i], a[i:])
x = sum(a[:i])
y = sum(a[i:])
min_v = min(min_v, abs(x - y))
print(min_v)
if __name__ == '__main__':
main()
|
s286247520
|
Accepted
| 160
| 24,492
| 643
|
def main():
N = int(input())
A = list(map(int, input().split()))
solve(N,A)
def solve(N,A):
ans = float("inf")
# ans = TLE(N,A,ans)
ans = AC(N,A,ans)
print(ans)
def TLE(N,A, ans):
for i in range(1, N):
snk = sum(A[:i])
ari = sum(A[i:])
ans = min(ans, abs(snk - ari))
return ans
def AC(N,A,ans):
cum_A = [None] * N
cum_A[0] = A[0]
for i in range(1, N):
cum_A[i] = cum_A[i-1] + A[i]
for i in range(N-1):
snk = cum_A[i]
ari = cum_A[-1] - snk
ans = min(ans, abs(snk - ari))
return ans
if __name__ == "__main__":
main()
|
s412692028
|
p02972
|
u366886346
| 2,000
| 1,048,576
|
Wrong Answer
| 593
| 13,900
| 241
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
n=int(input())
a=list(map(int,input().split()))
ans=[0]*n
for i in range(n):
num1=n-i
num2=0
for j in range((n//num1)-1):
num2+=ans[num1*(j+2)-1]
num2%=2
ans[-1-i]=(num2+a[-1-i])%2
print(ans.count(1))
print(*ans)
|
s143299754
|
Accepted
| 597
| 17,804
| 333
|
n=int(input())
a=list(map(int,input().split()))
ans=[0]*n
for i in range(n):
num1=n-i
num2=0
for j in range((n//num1)-1):
num2+=ans[num1*(j+2)-1]
num2%=2
ans[-1-i]=(num2+a[-1-i])%2
print(ans.count(1))
ans2=[]
for i in range(n):
if ans[i]==1:
ans2.append(i+1)
if len(ans2)!=0:
print(*ans2)
|
s941772451
|
p03386
|
u729938879
| 2,000
| 262,144
|
Wrong Answer
| 2,242
| 4,212
| 129
|
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())
num = [i for i in range(a, b+1)]
ans= set(num[:k]).union(num[-k:])
for i in ans:
print(i)
|
s586355860
|
Accepted
| 18
| 3,060
| 228
|
a, b, k = map(int, input().split())
length = b+1 -a
num = []
for i in range(a, min(b, a+k-1)+1):
print(i)
num.append(i)
for i in range(max(a, b - (k-1)), b+1):
if i not in num:
num.append(i)
print(i)
|
s763832908
|
p04031
|
u188827677
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,064
| 175
|
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
n = int(input())
a = list(map(int, input().split()))
m = sum(a)/n
if int(m)+1 - m < m - int(m): m = int(m)+1
else: int(m)
ans = 0
for i in a: ans += abs(i - m)**2
print(ans)
|
s706301903
|
Accepted
| 29
| 9,028
| 183
|
n = int(input())
a = list(map(int, input().split()))
m = sum(a)//n
ans = float("inf")
for i in range(m, m+2):
t = 0
for j in a:
t += abs(j-i)**2
ans = min(t,ans)
print(ans)
|
s581367115
|
p03377
|
u305965165
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 103
|
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 = [int(i) for i in input().split()]
if a<=x and x<=a+b:
print("Yes")
else:
print("No")
|
s996332439
|
Accepted
| 18
| 2,940
| 103
|
a, b, x = [int(i) for i in input().split()]
if a<=x and x<=a+b:
print("YES")
else:
print("NO")
|
s928406837
|
p03434
|
u992076031
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 240
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
N = int(input())
alist = input().split()
alist = sorted(alist, reverse=True)
print(alist)
count = 1
alice = 0
bob = 0
for a in alist:
if count % 2:
alice += int(a)
else:
bob += int(a)
count+=1
print(alice - bob)
|
s960353920
|
Accepted
| 18
| 3,060
| 252
|
N = int(input())
alist = input().split()
aint = list(map(int, alist))
aint = sorted(aint, reverse=True)
count = 1
alice = 0
bob = 0
for a in aint:
if count % 2:
alice += a
else:
bob += a
count = count + 1
print(alice - bob)
|
s118831430
|
p03721
|
u503901534
| 2,000
| 262,144
|
Wrong Answer
| 293
| 3,060
| 185
|
There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3.
|
n, k = map(int,input().split())
aa = 0
for i in range(n):
a,b = map(int,input().split())
if k -b < 0:
None
else:
aa = a
k = k - b
print(aa)
|
s979746519
|
Accepted
| 392
| 13,100
| 383
|
n, k = map(int,input().split())
dcic = {}
aa = set([])
for i in range(n):
a,b = map(int,input().split())
aa.add(a)
if a in dcic:
dcic[a] = dcic[a] + b
else:
dcic.update({a:b})
bb = list(sorted(list(aa)))
cc = 0
s = 0
for i in range(len(bb)):
if s < k:
s = s + dcic[bb[i]]
cc = bb[i]
else:
ccc = bb[i]
print(cc)
|
s801875166
|
p03166
|
u479719434
| 2,000
| 1,048,576
|
Wrong Answer
| 500
| 23,936
| 920
|
There is a directed graph G with N vertices and M edges. The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i. G **does not contain directed cycles**. Find the length of the longest directed path in G. Here, the length of a directed path is the number of edges in it.
|
from collections import deque
def main():
N, M = map(int, input().split())
e = [[] for _ in range(N)]
input_edges = [0]*N
for i in range(M):
u, v = map(int, input().split())
u -= 1
v -= 1
e[u].append(v)
input_edges[v] += 1
queue = deque()
for i, input_edge in enumerate(input_edges):
if input_edge == 0:
queue.appendleft(i)
topological_sorted = []
while queue:
v = queue.pop()
topological_sorted.append(v)
for other_side in e[v]:
input_edges[other_side] -= 1
if input_edges[other_side] == 0:
queue.appendleft(other_side)
dp = [10 ** 9] * N
dp[topological_sorted[0]] = 0
for v in topological_sorted:
for other_side in e[v]:
dp[other_side] = min(dp[v] + 1, dp[other_side])
print(max(dp))
if __name__ == "__main__":
main()
|
s088394933
|
Accepted
| 584
| 59,008
| 580
|
import sys
sys.setrecursionlimit(10**6)
def dfs(v):
global dp
if dp[v] != -1:
return dp[v]
res = 0
for next_v in e[v]:
res = max(dfs(next_v) + 1, res)
dp[v] = res
return res
def main():
N, M = map(int, input().split())
global e
e = [[] for _ in range(N)]
for i in range(M):
u, v = map(int, input().split())
u -= 1
v -= 1
e[u].append(v)
global dp
dp = [-1]*N
ans = -1
for v in range(N):
ans = max(ans, dfs(v))
print(ans)
if __name__ == "__main__":
main()
|
s663577792
|
p03162
|
u652150585
| 2,000
| 1,048,576
|
Wrong Answer
| 611
| 33,908
| 221
|
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
n=int(input())
l=[list(map(int,input().split())) for _ in range(n)]
dp=[0]*3
for i in range(n):
dp[0],dp[1],dp[2]=max(dp[1],dp[2])+l[i][0],max(dp[0],dp[2])+l[i][1],max(dp[0],dp[1])+l[i][2]
print(dp)
print(max(dp))
|
s796168522
|
Accepted
| 473
| 30,580
| 222
|
n=int(input())
l=[list(map(int,input().split())) for _ in range(n)]
dp=[0]*3
for i in range(n):
dp[0],dp[1],dp[2]=max(dp[1],dp[2])+l[i][0],max(dp[0],dp[2])+l[i][1],max(dp[0],dp[1])+l[i][2]
#print(dp)
print(max(dp))
|
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