message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a bookshelf which can fit n books. The i-th position of bookshelf is a_i = 1 if there is a book on this position and a_i = 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
In one move, you can choose some contiguous segment [l; r] consisting of books (i.e. for each i from l to r the condition a_i = 1 holds) and:
* Shift it to the right by 1: move the book at index i to i + 1 for all l β€ i β€ r. This move can be done only if r+1 β€ n and there is no book at the position r+1.
* Shift it to the left by 1: move the book at index i to i-1 for all l β€ i β€ r. This move can be done only if l-1 β₯ 1 and there is no book at the position l-1.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
For example, for a = [0, 0, 1, 0, 1] there is a gap between books (a_4 = 0 when a_3 = 1 and a_5 = 1), for a = [1, 1, 0] there are no gaps between books and for a = [0, 0,0] there are also no gaps between books.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 β€ t β€ 200) β the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 β€ n β€ 50) β the number of places on a bookshelf. The second line of the test case contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 1), where a_i is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.
Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).
Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3; 3] to the right and the segment [4; 5] to the right. After all moves, the books form the contiguous segment [5; 7]. So the answer is 2.
In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.
In the third test case of the example, you can shift the segment [5; 5] to the left and then the segment [4; 4] to the left again. After all moves, the books form the contiguous segment [1; 3]. So the answer is 2.
In the fourth test case of the example, you can shift the segment [1; 1] to the right, the segment [2; 2] to the right, the segment [6; 6] to the left and then the segment [5; 5] to the left. After all moves, the books form the contiguous segment [3; 4]. So the answer is 4.
In the fifth test case of the example, you can shift the segment [1; 2] to the right. After all moves, the books form the contiguous segment [2; 5]. So the answer is 1.
Submitted Solution:
```
for _ in range(int(input())):
x=int(input())
ll=input().split(" ")
start=9999999
zero=0
ends=0
if ll.count("1")==1:
zero=0
else:
for i in range(0,len(ll)):
if start==9999999 and ll[i]=='1':
start=i
end=i
if start!=9999999 and ll[i]!='1':
zero+=1
if ll[i]=="1":
ends=i
print(zero)
``` | instruction | 0 | 38,868 | 8 | 77,736 |
No | output | 1 | 38,868 | 8 | 77,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | instruction | 0 | 39,350 | 8 | 78,700 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
def main():
n, h = map(int, input().split())
l = -1
r = 10 ** 18
while (r - l != 1):
m = (r + l) // 2
t = m + h - 1
if (m > h):
if (t % 2 == 1):
j = t // 2 + 1 - h + 1
s = j * (j + 1) // 2 + j * (h - 1) + (m - j) * (m - j + 1) // 2
if (s >= n):
r = m
else:
l = m
else:
j = t // 2 - h + 1
s = j * (j + 1) // 2 + j * (h - 1) + (m - j) * (m - j + 1) // 2
if (s >= n):
r = m
else:
l = m
else:
s = m * (m + 1) // 2
if (s >= n):
r = m
else:
l = m
print(r)
main()
``` | output | 1 | 39,350 | 8 | 78,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | instruction | 0 | 39,351 | 8 | 78,702 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
import math
def sum(x):
return (x*(x+1))//2
def check(x):
ans = 0
if x <= h:
ans = sum(x)
else:
md = (x - h + 2)/2.0
if md - math.floor(md) == 0:
ans = sum(h+md-1) - sum(h-1) + sum(x-md)
else:
md = math.floor(md)
ans = sum(h+md-1) - sum(h-1) + sum(h+md-1)
if ans<n:
return 1
else:
return 0
n, h = input().strip().split()
n = (int)(n)
h = (int)(h)
l = 0
r = n
while l<r :
mid = (l+r)//2
if check(mid) != 0:
l = mid+1
else:
r = mid
if check(r) == 0:
ans1 = r
else:
ans1 = r + 1
print(ans1)
``` | output | 1 | 39,351 | 8 | 78,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | instruction | 0 | 39,352 | 8 | 78,704 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
n, H = map(int, input().split())
def f(x):
if x <= H:
return x * (x + 1) // 2
if (x + H) % 2 == 0:
M = (x + H) // 2
return M * M - (H * (H - 1) // 2)
else:
M = (x + H - 1) // 2
return M * (M + 1) - H * (H - 1) // 2
lo, hi = -1, n + 1
while hi - lo > 1:
mid = (lo + hi) // 2
if f(mid) >= n:
hi = mid
else:
lo = mid
print(hi)
``` | output | 1 | 39,352 | 8 | 78,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | instruction | 0 | 39,353 | 8 | 78,706 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
def sumab(a, b):
return (a+b)*(b-a+1)//2
def calc(spot, H):
if spot <= H: return spot * (spot + 1) // 2
if (H - spot) % 2 == 0: return sumab(1,H+(spot-H)//2)+sumab(H,H+(spot-H)//2-1)
return sumab(1,H+(spot-H)//2)+sumab(H,H+(spot-H)//2-1)+H+(spot-H)//2
def solve():
n,H = map(int,input().split())
if n == 1:
print(1)
return 0
st = 1
en = 10**20
while st < en:
mid = (st+en)//2
if calc(mid,H) >= n:
en = mid
else:
st = mid+1
print(st)
solve()
``` | output | 1 | 39,353 | 8 | 78,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | instruction | 0 | 39,354 | 8 | 78,708 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
from math import *
n, h = (int(x) for x in input().split(' '))
if (h + 1) * h // 2 >= n:
a = int(ceil((-1 + sqrt(1 + 8 * n)) / 2))
if (a + 1) * a // 2 >= n:
print(a)
else:
print(a + 1)
else:
n -= int((h + 1) * h / 2)
n -= h
if n > 0:
a = int((-2 * h - 1 + sqrt((2 * h + 1) * (2 * h + 1) + 4 * n)) / 2)
if (2 * h + 1 + a) * a >= n:
print(a * 2 + h + 1)
elif (2 * h + 1 + a) * a + h + a + 1 >= n:
print(a * 2 + 1 + h + 1)
else:
print(a * 2 + 2 + h + 1)
else:
print(h + 1)
``` | output | 1 | 39,354 | 8 | 78,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | instruction | 0 | 39,355 | 8 | 78,710 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
n, h = [int(x) for x in input().split()]
l = -1
r = n + 1
m = (l + r) // 2
while r - l > 1:
if m <= h:
v = m * (m + 1) // 2
else:
v = m * (m + 1) // 2 - h * (h - 1) // 2 + m * (m - 1) // 2
if v <= n:
l = m
else:
r = m
m = (l + r) // 2
if l < h:
v = l * (l + 1) // 2
print(l + (n - v) // l + ((n - v) % l != 0))
exit()
else:
v = l * (l + 1) // 2 - h * (h - 1) // 2 + l * (l - 1) // 2
print(l - h + l + (n - v) // l + ((n - v) % l != 0))
``` | output | 1 | 39,355 | 8 | 78,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | instruction | 0 | 39,356 | 8 | 78,712 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
import math
# python math.sqrt(ζζ―) δΊεεΌζΉ
def g(a,b):
return (a+b) * (b-a+1)/2
def f(a,b):
return g(a,b) + g(1,b)
n, h = map(int, input().strip().split())
if(h * (1 + h) / 2 >= n):
l = int(1)
r = int(h)
while(r > l):
mid = int((l+r) /2)
if(mid * (1+mid) < n * 2):
l = mid + 1
else:
r = mid
print(int(l))
else:
l = h
r = 1500000000
while(r > l):
mid = int((l+r) / 2)
if(f(h,mid) < n):
l = mid + 1
else:
r = mid
if(n <= g(h,l)+g(1,l-1)):
print(int((l-h) + l))
else:
print(int(l-h+l+1))
``` | output | 1 | 39,356 | 8 | 78,713 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...] | instruction | 0 | 39,357 | 8 | 78,714 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
def getSum1(x):
return (x + 1) * x // 2;
def getSum(l, r):
return getSum1(r) - getSum1(l - 1)
n, h = map(int, input().split())
l = 1
r = n
while l != r:
m = (l + r + 1) >> 1
cnt = 0
if m > h:
cnt += getSum(h, m - 1)
cnt += getSum1(m)
if cnt > n:
r = m - 1
else: l = m
cnt = 0
ans = 0
if l > h:
cnt += getSum(h, l - 1)
ans += l - h
cnt += getSum1(l)
ans += l
n -= cnt
ans += n // l
if n % l != 0:
ans += 1
print(ans)
``` | output | 1 | 39,357 | 8 | 78,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]
Submitted Solution:
```
n, H = map(int, input().split())
def can(x):
if x <= H:
return n <= (1 + x) * x // 2
L = x + H - 1
h = L // 2
s = (1 + h) * h
if L % 2 == 1:
s += h + 1
s -= H * (H - 1) // 2
return n <= s
m, M = 0, int(1e18) + 5
while M - m > 1:
mid = m + (M - m) // 2
if can(mid):
M = mid
else:
m = mid
print(M)
``` | instruction | 0 | 39,358 | 8 | 78,716 |
Yes | output | 1 | 39,358 | 8 | 78,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]
Submitted Solution:
```
n,h = map(int,input().split())
left = 0
right = 10**18
while (left +1 != right):
mid = (left + right) // 2
if ((mid + min(mid,h))%2 == 0):
ans = ((mid + min(h,mid)) // 2)*((mid+min(mid,h))//2) - min(mid,h)*(min(mid,h)-1)//2
if (ans >=n):
right = mid
else:
left = mid
else:
ans = ((mid + min(mid,h)) // 2 + 1) * ((mid + min(mid,h))//2) - min(mid,h)*(min(mid,h)-1)//2
if (ans >=n):
right = mid
else:
left = mid
print(right)
``` | instruction | 0 | 39,359 | 8 | 78,718 |
Yes | output | 1 | 39,359 | 8 | 78,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]
Submitted Solution:
```
def max_number_of_packs(number_of_spots, fence_height):
if number_of_spots > fence_height:
k = (number_of_spots - fence_height) // 2 + fence_height
l = number_of_spots - k
return (k * (k+1) // 2) + (l * (l+1) // 2) + (l * (fence_height-1))
else:
return number_of_spots * (number_of_spots+1) // 2
def binary_search(left, right, number_of_packs, fence_height):
if left >= right - 1:
return right
middle = (left + right) // 2
max_packs = max_number_of_packs(middle, fence_height)
if max_packs < number_of_packs:
return binary_search(middle, right, number_of_packs, fence_height)
else:
return binary_search(left, middle, number_of_packs, fence_height)
n, h = map(int, input().split())
print(binary_search(0, 10**18, n, h))
``` | instruction | 0 | 39,360 | 8 | 78,720 |
Yes | output | 1 | 39,360 | 8 | 78,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]
Submitted Solution:
```
def binsearch():
l = 0
r = 10**10
while r - l > 1:
m = (l + r) // 2
if ((m * m + m) // 2 <= n):
l = m
else:
r = m
return l
def solve(x):
val = x * x + (h - h * h) // 2
if val > n or x < 1:
return 2 * 10 ** 18
else:
return (2 * x - h + (0 if n == val else (n - val - 1) // x + 1))
n, h = map(int, input().split())
if (h + h * h) / 2 > n:
x = binsearch()
val = (x * x + x) // 2
if val == n:
print(x)
elif n - val <= x:
print(x + 1)
elif n - val <= 2 * x:
print(x + 2)
else:
print(x + 3)
else:
x = int((n + (h * h - h) // 2)**0.5)
ans = 2 * 10**18
for c in range(x - 100, x + 1):
ans = min(ans, solve(c))
print(ans)
``` | instruction | 0 | 39,361 | 8 | 78,722 |
Yes | output | 1 | 39,361 | 8 | 78,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]
Submitted Solution:
```
def S (l):
return l * (l + 1) // 2
def f (l, h):
if l <= h:
return S(l)
s = S(h)
l -= h
s += l * h
l -= 1
s += S(l // 2) + S(l // 2 - (1 - l % 2))
return s
n, H = map(int, input().split())
l = 0
r = n + 1
while r - l > 1:
md = (l + r) // 2
if f(md, H) >= n:
r = md
else:
l = md
print(r)
``` | instruction | 0 | 39,362 | 8 | 78,724 |
No | output | 1 | 39,362 | 8 | 78,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]
Submitted Solution:
```
(n, H) = [int(i) for i in input().split()]
start = 0
end = n
while start != end - 1:
mid = (start + end) // 2
h = min(H, mid)
result = mid ** 2 - (h * (h - 1)) // 2
if result <= n:
start = mid
else:
end = mid
if start <= H:
length = start
M = length * (length + 1) // 2
else:
length = start + abs(start - H)
M = length * (length + 1) // 2 + ((start - 1 + H) * (start - H)) // 2
tomake = (n - M) // start
if (n - M) % start != 0:
tomake += 1
ans = length + tomake
print(ans)
``` | instruction | 0 | 39,363 | 8 | 78,726 |
No | output | 1 | 39,363 | 8 | 78,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]
Submitted Solution:
```
import math
n,h=(map(int,input().strip().split(' ')))
x=(h*(h+1))//2
if(x==n):
print(h)
elif(h==1):
print(n)
elif(n==1):
print(1)
elif(n<x):
low = 1
high = h
ans=10**20
while(low<=high):
mid=(low+high)//2
temp=(mid*(mid+1))//2
if(temp==n):
ans=mid
break
if(temp<n):
low=mid+1
else:
ans=min(ans,mid)
high=mid-1
print(ans)
else:
low = 1
high = 10**10
ans = -10**20
while(low<=high):
mid=(low+high)//2
temp=((mid-1)*(mid))-((h*(h+1))//2)+h+mid
# print(low,high,temp)
if(temp==n):
ans=mid
break
if(temp<n):
low=mid+1
ans=max(ans,mid)
else:
high=mid-1
# print(ans)
print(ans+max(ans-h,1)+1)
``` | instruction | 0 | 39,364 | 8 | 78,728 |
No | output | 1 | 39,364 | 8 | 78,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.
Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.
Finally you ended up with the following conditions to building the castle:
* h1 β€ H: no sand from the leftmost spot should go over the fence;
* For any <image> |hi - hi + 1| β€ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
* <image>: you want to spend all the sand you brought with you.
As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.
Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.
Input
The only line contains two integer numbers n and H (1 β€ n, H β€ 1018) β the number of sand packs you have and the height of the fence, respectively.
Output
Print the minimum number of spots you can occupy so the all the castle building conditions hold.
Examples
Input
5 2
Output
3
Input
6 8
Output
3
Note
Here are the heights of some valid castles:
* n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
* n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)
The first list for both cases is the optimal answer, 3 spots are occupied in them.
And here are some invalid ones:
* n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
* n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]
Submitted Solution:
```
def calc(spot, H):
if spot <= H: return spot * (spot + 1) // 2
else: return (spot-H)*H + H * (H + 1) // 2
def solve():
n,H = map(int,input().split())
if n == 1:
print(1)
return 0
st = 1
en = 10**10
while st < en:
mid = (st+en)//2
if calc(mid,H) >= n:
en = mid
else:
st = mid+1
print(st)
solve()
``` | instruction | 0 | 39,365 | 8 | 78,730 |
No | output | 1 | 39,365 | 8 | 78,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly. | instruction | 0 | 39,688 | 8 | 79,376 |
Tags: dp, greedy
Correct Solution:
```
T = int(input())
while T>0:
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
pos = 0
for i in range(n-1):
if abs(a[i] - a[i+1]) <= k:
pos += 1
d = max(0, a[i+1] - k)
m += (a[i] - d)
else:
if a[i] < a[i+1]:
d = max(0, a[i+1] - k)
m -= (d - a[i])
if m < 0:
break
else:
pos += 1
else:
d = max(0, a[i+1] - k)
m += (a[i] - d)
pos += 1
T -= 1
print("YES" if pos == n-1 else "NO")
``` | output | 1 | 39,688 | 8 | 79,377 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly. | instruction | 0 | 39,689 | 8 | 79,378 |
Tags: dp, greedy
Correct Solution:
```
for nt in range(int(input())):
n,m,k = map(int,input().split())
a = list(map(int,input().split()))
ans = "YES"
for i in range(1,n):
if a[i]<=a[i-1]:
m += min(a[i-1],(a[i-1]-a[i])+k)
elif a[i-1]>=a[i]-k:
m += min(a[i-1],a[i-1]-(a[i]-k))
else:
m -= (a[i]-k-a[i-1])
if m<0:
ans = "NO"
break
print (ans)
``` | output | 1 | 39,689 | 8 | 79,379 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly. | instruction | 0 | 39,690 | 8 | 79,380 |
Tags: dp, greedy
Correct Solution:
```
import sys
def input():
return sys.stdin.readline()[:-1]
T = int(input())
for i in range(T):
flag = True
n,m,k = list(map(int,input().split()))
h = list(map(int,input().split()))
for i in range(n-1):
x=max(h[i+1]-k,0)
if x-h[i]>m:
flag = False
break
else:
m -= x-h[i]
if flag:
print("YES")
else:
print("NO")
``` | output | 1 | 39,690 | 8 | 79,381 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly. | instruction | 0 | 39,691 | 8 | 79,382 |
Tags: dp, greedy
Correct Solution:
```
t=int(input())
for i in range(t):
[n,m,k]=[int(x) for x in input().split()]
h=[int(x) for x in input().split()]
flag=True
for j in range(0,n-1):
if h[j]>h[j+1]:
if h[j+1]>=k:
m+=(k+h[j]-h[j+1])
else:
m+=(h[j])
elif h[j]<=h[j+1]:
if h[j+1]-h[j]<=k:
m+=((k-(h[j+1]-h[j])) if h[j+1]>=k else (h[j]))
else:
if m>=(h[j+1]-h[j]-k):
m-=(h[j+1]-h[j]-k)
else:
flag=False
break
if flag:
print("YES")
else:
print("NO")
``` | output | 1 | 39,691 | 8 | 79,383 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly. | instruction | 0 | 39,692 | 8 | 79,384 |
Tags: dp, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n,m,k = map(int,input().split())
h = list(map(int,input().split()))
now = h[0]
t = True
for i in range(1,n):
if now < h[i]:
if now + m + k < h[i]:
t = False
break
else:
if now <= h[i]-k:
m -= h[i] - k - now
else:
m += now - max(h[i]-k,0)
else:
m += now - max(h[i]-k,0)
now = h[i]
if t:
print('YES')
else:
print('NO')
``` | output | 1 | 39,692 | 8 | 79,385 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly. | instruction | 0 | 39,693 | 8 | 79,386 |
Tags: dp, greedy
Correct Solution:
```
def can( n, m, k, h ):
for i in range( n - 1 ):
if m + h[ i ] + k < h[ i + 1 ]:
return False
if h[ i ] + k < h[ i + 1 ]:
m -= h[ i + 1 ] - h[ i ] - k
else:
m += h[ i ] - max( 0, h[ i + 1 ] - k )
return True
t = int( input() )
for tc in range(t):
n, m, k = map( int, input().split() )
h = list( map( int, input().split() ) )
print( "YES" if can( n, m, k, h ) else "NO" )
``` | output | 1 | 39,693 | 8 | 79,387 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly. | instruction | 0 | 39,694 | 8 | 79,388 |
Tags: dp, greedy
Correct Solution:
```
for i in range(int(input())):
n,m,k = map(int,input().split())
s = [int(i) for i in input().split()]
c = 1
for i in range(n-1):
if s[i]>=s[i+1]:
m+=min(s[i]-s[i+1]+k,s[i])
else:
if s[i+1]-s[i]>k:
m-=s[i+1]-k-s[i]
else:
m+=min(k-(s[i+1]-s[i]),s[i])
if m<0:
c = 0
break
if c:
print("YES")
else:
print("NO")
``` | output | 1 | 39,694 | 8 | 79,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly. | instruction | 0 | 39,695 | 8 | 79,390 |
Tags: dp, greedy
Correct Solution:
```
# kartikay26
from math import *
from collections import *
from itertools import *
from functools import *
from random import *
def getl(t=int): return [t(x) for x in input().split()]
def get(t=int): return t(input())
alphabet = [chr(x) for x in range(ord('a'), ord('z')+1)]
alnum = lambda x: ord(x) - ord('a')
def main():
for _ in range(get()):
if test():
print("YES")
else:
print("NO")
def test():
n,m,k = getl()
a = getl()
for i in range(n-1):
# print("dbg",i,m,a)
if m >= a[i+1] - a[i] - k:
m -= (a[i+1] - a[i] - k)
a[i] += (a[i+1] - a[i] - k)
if a[i] < 0:
m += a[i]
a[i] = 0
else:
return False
return True
if __name__ == "__main__":
main()
``` | output | 1 | 39,695 | 8 | 79,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly.
Submitted Solution:
```
for i in range(int(input())):
n, m, k = [int(i) for i in input().split()]
'''
n - ΠΊΠΎΠ»ΠΈΡΠ΅ΡΡΠ²ΠΎ ΡΡΠΎΠ»Π±ΡΠΎΠ²
m - ΠΊΠΎΠ»ΠΈΡΠ΅ΡΡΠ²ΠΎ ΠΊΡΠ±ΠΈΠΊΠΎΠ² Π² ΡΡΠΊΠ·Π°ΠΊΠ΅ ΠΈΠ·Π½Π°ΡΠ°Π»ΡΠ½ΠΎ
k - ΠΌΠ°ΠΊΡ ΠΌΠΎΠ΄ΡΠ»Ρ ΡΠ°Π·Π½ΠΎΡΡΠΈ
'''
a = [int(i) for i in input().split()]
if n == 1:
print("YES")
else:
for j in range(n - 1):
mindelta = a[j + 1] - (a[j] + k)
if a[j] + k >= a[j + 1]:
m += min(a[j] + k - a[j + 1], a[j])
else:
if a[j] + k + m >= a[j + 1]:
m -= a[j + 1] - a[j] - k
else:
print("NO")
break
else:
print("YES")
``` | instruction | 0 | 39,696 | 8 | 79,392 |
Yes | output | 1 | 39,696 | 8 | 79,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly.
Submitted Solution:
```
t=int(input())
while t>0:
t-=1
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
x=[]
if n==1:
print('YES')
else:
q=1
for i in range(n-1):
z=a[i]-a[i+1]
#print(z)
if z>=0:
m+=z
m+=min(k,a[i]-z)
q=1
else:
if abs(z)>k:
y=a[i+1]-k-a[i]
if m>=y:
m=m-y
else:
q=0
break
else:
y=a[i]-(a[i+1]-k)
m+=min(y,a[i])
#print(m)
'''if abs(z+k)<=m and :
q=1
m=m-abs(z+k)
else:
q=0
break'''
if q==0:
print('NO')
else:
print('YES')
'''if len(x)==1:
if x[0]==k:
print('YES')
else:
print('NO')
else:
q=0
for i in range(len(x)-1):
z=abs(x[i]+x[i+1])
if z!=k:
q=1
break
if q==0:
print('YES')
else:
print('NO')'''
'''q=1
for i in range(n-1):
z=abs(a[i]-a[i-1])
if z>k:'''
``` | instruction | 0 | 39,697 | 8 | 79,394 |
Yes | output | 1 | 39,697 | 8 | 79,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly.
Submitted Solution:
```
import sys
n = int(sys.stdin.readline())
for i in range(n):
res = 1
prelim = sys.stdin.readline().split()
col_num = int(prelim[0])
block_in_beg = int(prelim[1])
climb_diff = int(prelim[2])
block_heights = sys.stdin.readline().split()
for j in range(col_num):
block_heights[j] = int(block_heights[j])
for j in range(col_num-1):
height_diff = block_heights[j] - block_heights[j+1]
if(height_diff< -1*climb_diff):
if(height_diff+block_in_beg < -1*climb_diff):
print("NO")
res = 0
break
else:
block_in_beg += (height_diff + climb_diff)
else:
tmp_block_gain = (height_diff + climb_diff)
if tmp_block_gain > block_heights[j]:
block_in_beg += block_heights[j]
else:
block_in_beg += tmp_block_gain
if res:
print("YES")
``` | instruction | 0 | 39,698 | 8 | 79,396 |
Yes | output | 1 | 39,698 | 8 | 79,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly.
Submitted Solution:
```
for _ in range(int(input())):
n,m,k = map(int,input().split())
Height = [int(x) for x in input().split()]
flag = 0
for i in range(len(Height)-1):
if(Height[i+1]-Height[i]>0):
if(Height[i+1]-Height[i]<=k):
m+=min(Height[i],Height[i]-(Height[i+1]-k))
else:
Needed = Height[i+1]-k - Height[i]
if(m>=Needed):
m-=Needed
else:
print("NO")
flag = 1
break
elif(Height[i+1]==Height[i]):
m+=min(k,Height[i])
else:
m+=min(Height[i],Height[i]-(Height[i+1]-k))
if(flag == 0):
print("YES")
``` | instruction | 0 | 39,699 | 8 | 79,398 |
Yes | output | 1 | 39,699 | 8 | 79,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly.
Submitted Solution:
```
from sys import stdin
def ii(): return int(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def li(): return list(mi())
for _ in range(ii()):
p=0
n, m, k=mi()
a=li()
for i in range(n-1):
x=max(0, a[i+1]-m)
if a[i]<x:
k-=x-a[i]
if k<0:
p=1
break
else:
k+=a[i]-x
print(['YES', 'NO'][p])
``` | instruction | 0 | 39,700 | 8 | 79,400 |
No | output | 1 | 39,700 | 8 | 79,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly.
Submitted Solution:
```
from sys import stdin
input=stdin.readline
for i in range(int(input())):
n,m,k=map(int,input().split())
h=list(map(int,input().split()))
bag=m
for i in range(n-1):
if h[i]>h[i+1]:
bag+=h[i]
h[i]=0
if h[i]<=h[i+1]:
dif=h[i+1]-h[i]
if k>dif:
bag+=k-dif
elif dif-k<=bag:
bag-=dif-k
else:
print('NO')
break
else:
print('YES')
``` | instruction | 0 | 39,701 | 8 | 79,402 |
No | output | 1 | 39,701 | 8 | 79,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly.
Submitted Solution:
```
inp = lambda: map(int, input().rstrip().split())
for i in range(int(input())):
n, m, k = inp()
h = list(inp())
i = 0
while i < n - 1 and m >= 0:
x = h[i]
y = h[i + 1]
if x >= y:
m += (x - y + k)
i += 1
elif y > x and y - k - x > 0:
diff = y - k - x
if diff > m:
m = -1
else:
m -= diff
i += 1
elif y > x:
m += x - y + k
i += 1
if m >= 0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 39,702 | 8 | 79,404 |
No | output | 1 | 39,702 | 8 | 79,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is playing a video game called Block Adventure. In Block Adventure, there are n columns of blocks in a row, and the columns are numbered from 1 to n. All blocks have equal heights. The height of the i-th column is represented as h_i, which is the number of blocks stacked in the i-th column.
Gildong plays the game as a character that can stand only on the top of the columns. At the beginning, the character is standing on the top of the 1-st column. The goal of the game is to move the character to the top of the n-th column.
The character also has a bag that can hold infinitely many blocks. When the character is on the top of the i-th column, Gildong can take one of the following three actions as many times as he wants:
* if there is at least one block on the column, remove one block from the top of the i-th column and put it in the bag;
* if there is at least one block in the bag, take one block out of the bag and place it on the top of the i-th column;
* if i < n and |h_i - h_{i+1}| β€ k, move the character to the top of the i+1-st column. k is a non-negative integer given at the beginning of the game. Note that it is only possible to move to the next column.
In actions of the first two types the character remains in the i-th column, and the value h_i changes.
The character initially has m blocks in the bag. Gildong wants to know if it is possible to win the game. Help Gildong find the answer to his question.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 1000). Description of the test cases follows.
The first line of each test case contains three integers n, m, and k (1 β€ n β€ 100, 0 β€ m β€ 10^6, 0 β€ k β€ 10^6) β the number of columns in the game, the number of blocks in the character's bag at the beginning, and the non-negative integer k described in the statement.
The second line of each test case contains n integers. The i-th integer is h_i (0 β€ h_i β€ 10^6), the initial height of the i-th column.
Output
For each test case, print "YES" if it is possible to win the game. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
5
3 0 1
4 3 5
3 1 2
1 4 7
4 10 0
10 20 10 20
2 5 5
0 11
1 9 9
99
Output
YES
NO
YES
NO
YES
Note
In the first case, Gildong can take one block from the 1-st column, move to the 2-nd column, put the block on the 2-nd column, then move to the 3-rd column.
In the second case, Gildong has to put the block in his bag on the 1-st column to get to the 2-nd column. But it is impossible to get to the 3-rd column because |h_2 - h_3| = 3 > k and there is no way to decrease the gap.
In the fifth case, the character is already on the n-th column from the start so the game is won instantly.
Submitted Solution:
```
for i in range(int(input())):
n, m,k=map(int,input().split())
l=list(map(int,input().split()))
sivi=0
for i in range(n-1):
if l[i]<l[i+1]:
if abs(l[i]-l[i+1])>k:
flag=0
while m>0:
l[i]+=1
m-=1
if abs(l[i]-l[i+1])==k:
flag=1
break
if flag==0:
sivi=1
break
elif l[i]>l[i+1]:
e=(l[i]-l[i+1])
m+=(e+k)
if sivi==0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 39,703 | 8 | 79,406 |
No | output | 1 | 39,703 | 8 | 79,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Numbers 1, 2, 3, ... n (each integer from 1 to n once) are written on a board. In one operation you can erase any two numbers a and b from the board and write one integer (a + b)/(2) rounded up instead.
You should perform the given operation n - 1 times and make the resulting number that will be left on the board as small as possible.
For example, if n = 4, the following course of action is optimal:
1. choose a = 4 and b = 2, so the new number is 3, and the whiteboard contains [1, 3, 3];
2. choose a = 3 and b = 3, so the new number is 3, and the whiteboard contains [1, 3];
3. choose a = 1 and b = 3, so the new number is 2, and the whiteboard contains [2].
It's easy to see that after n - 1 operations, there will be left only one number. Your goal is to minimize it.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The only line of each test case contains one integer n (2 β€ n β€ 2 β
10^5) β the number of integers written on the board initially.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, in the first line, print the minimum possible number left on the board after n - 1 operations. Each of the next n - 1 lines should contain two integers β numbers a and b chosen and erased in each operation.
Example
Input
1
4
Output
2
2 4
3 3
3 1 | instruction | 0 | 39,802 | 8 | 79,604 |
Tags: constructive algorithms, data structures, greedy, implementation, math
Correct Solution:
```
#dt = {} for i in x: dt[i] = dt.get(i,0)+1
import sys;input = sys.stdin.readline
#import io,os; input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline #for pypy
inp,ip = lambda :int(input()),lambda :[int(w) for w in input().split()]
for _ in range(inp()):
n = inp()
prev = n
print(2)
for i in range(n-1,0,-1):
print(i,prev)
prev = (i+prev-1)//2 +1
``` | output | 1 | 39,802 | 8 | 79,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Numbers 1, 2, 3, ... n (each integer from 1 to n once) are written on a board. In one operation you can erase any two numbers a and b from the board and write one integer (a + b)/(2) rounded up instead.
You should perform the given operation n - 1 times and make the resulting number that will be left on the board as small as possible.
For example, if n = 4, the following course of action is optimal:
1. choose a = 4 and b = 2, so the new number is 3, and the whiteboard contains [1, 3, 3];
2. choose a = 3 and b = 3, so the new number is 3, and the whiteboard contains [1, 3];
3. choose a = 1 and b = 3, so the new number is 2, and the whiteboard contains [2].
It's easy to see that after n - 1 operations, there will be left only one number. Your goal is to minimize it.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The only line of each test case contains one integer n (2 β€ n β€ 2 β
10^5) β the number of integers written on the board initially.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, in the first line, print the minimum possible number left on the board after n - 1 operations. Each of the next n - 1 lines should contain two integers β numbers a and b chosen and erased in each operation.
Example
Input
1
4
Output
2
2 4
3 3
3 1 | instruction | 0 | 39,807 | 8 | 79,614 |
Tags: constructive algorithms, data structures, greedy, implementation, math
Correct Solution:
```
import math
for x in range(int(input())):
n=int(input())
a=[]
o=[]
e=[]
ans=[]
if n%2==0:
t=0
else:
t=1
for i in range(n):
if (i+1)%2==0:
e.append(i+1)
else:
o.append(i+1)
if n:
for i in range(n-1):
if t==0:
if len(e)>=2:
v1=e.pop()
v2=e.pop()
v3=(v1+v2)//2
ans.append((v1,v2))
if (v3)%2==0:
e.append(v3)
else:
o.append(v3)
elif len(o)>=2:
v1=o.pop()
v2=o.pop()
ans.append((v1,v2))
v3=(v1+v2)//2
if (v3)%2==0:
e.append(v3)
else:
o.append(v3)
else:
if len(e)>0:
v1=e.pop()
if len(o)>0:
v2=o.pop()
ans.append((v1,v2))
v3=math.ceil((v1+v2)/2)
if (v3)%2==0:
e.append(v3)
else:
o.append(v3)
else:
if len(o)>=2:
v1=o.pop()
v2=o.pop()
ans.append((v1,v2))
v3=(v1+v2)//2
if (v3)%2==0:
e.append(v3)
else:
o.append(v3)
elif len(e)>=2:
v1=e.pop()
v2=e.pop()
v3=(v1+v2)//2
ans.append((v1,v2))
if (v3)%2==0:
e.append(v3)
else:
o.append(v3)
else:
if len(e)>0:
v1=e.pop()
if len(o)>0:
v2=o.pop()
ans.append((v1,v2))
v3=math.ceil((v1+v2)/2)
if (v3)%2==0:
e.append(v3)
else:
o.append(v3)
if len(o)>=2 and len(e)>=2:
if o[-1]+o[-2]>e[-1]+e[-2]:
t=1
else:
t=0
if len(o)>0:
print(o[0])
else:
print(e[0])
for i in ans:
print(i[0],i[1])
``` | output | 1 | 39,807 | 8 | 79,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Numbers 1, 2, 3, ... n (each integer from 1 to n once) are written on a board. In one operation you can erase any two numbers a and b from the board and write one integer (a + b)/(2) rounded up instead.
You should perform the given operation n - 1 times and make the resulting number that will be left on the board as small as possible.
For example, if n = 4, the following course of action is optimal:
1. choose a = 4 and b = 2, so the new number is 3, and the whiteboard contains [1, 3, 3];
2. choose a = 3 and b = 3, so the new number is 3, and the whiteboard contains [1, 3];
3. choose a = 1 and b = 3, so the new number is 2, and the whiteboard contains [2].
It's easy to see that after n - 1 operations, there will be left only one number. Your goal is to minimize it.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The only line of each test case contains one integer n (2 β€ n β€ 2 β
10^5) β the number of integers written on the board initially.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, in the first line, print the minimum possible number left on the board after n - 1 operations. Each of the next n - 1 lines should contain two integers β numbers a and b chosen and erased in each operation.
Example
Input
1
4
Output
2
2 4
3 3
3 1 | instruction | 0 | 39,809 | 8 | 79,618 |
Tags: constructive algorithms, data structures, greedy, implementation, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n=int(input())
print(2)
if n==2:
print("1 2")
else:
print(str(n)+" "+str(n-2))
print(str(n-1)+" "+str(n-1))
j=1
for i in range(3,n):
print(str(n-j)+" "+str(n-(j+2)))
j+=1
``` | output | 1 | 39,809 | 8 | 79,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Numbers 1, 2, 3, ... n (each integer from 1 to n once) are written on a board. In one operation you can erase any two numbers a and b from the board and write one integer (a + b)/(2) rounded up instead.
You should perform the given operation n - 1 times and make the resulting number that will be left on the board as small as possible.
For example, if n = 4, the following course of action is optimal:
1. choose a = 4 and b = 2, so the new number is 3, and the whiteboard contains [1, 3, 3];
2. choose a = 3 and b = 3, so the new number is 3, and the whiteboard contains [1, 3];
3. choose a = 1 and b = 3, so the new number is 2, and the whiteboard contains [2].
It's easy to see that after n - 1 operations, there will be left only one number. Your goal is to minimize it.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The only line of each test case contains one integer n (2 β€ n β€ 2 β
10^5) β the number of integers written on the board initially.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, in the first line, print the minimum possible number left on the board after n - 1 operations. Each of the next n - 1 lines should contain two integers β numbers a and b chosen and erased in each operation.
Example
Input
1
4
Output
2
2 4
3 3
3 1
Submitted Solution:
```
from heapq import heappush as hpush
from heapq import heappop as hpop
from heapq import heapify
t = int(input())
for _ in range(t):
n = int(input())
if n == 1:
print(1)
elif n ==2:
print(2)
elif n>2:
arr = []
ans = []
for i in range(1,n):
if i!=(n-2):
val = -i
arr.append(val)
#print(arr)
temp = -(n + (n-2))//2
ans.append((n,n-2))
heapify(arr)
hpush(arr,temp)
#print(arr)
for i in range(n-2):
val1 = -hpop(arr)
val2 = -hpop(arr)
ans.append((val1,val2))
temp = -((val1+val2)//2)
hpush(arr,temp)
if len(arr)==1:
print(-arr[0])
length = len(ans)
for i in range(length):
print(f'{ans[i][0]} {ans[i][1]}')
``` | instruction | 0 | 39,814 | 8 | 79,628 |
No | output | 1 | 39,814 | 8 | 79,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Numbers 1, 2, 3, ... n (each integer from 1 to n once) are written on a board. In one operation you can erase any two numbers a and b from the board and write one integer (a + b)/(2) rounded up instead.
You should perform the given operation n - 1 times and make the resulting number that will be left on the board as small as possible.
For example, if n = 4, the following course of action is optimal:
1. choose a = 4 and b = 2, so the new number is 3, and the whiteboard contains [1, 3, 3];
2. choose a = 3 and b = 3, so the new number is 3, and the whiteboard contains [1, 3];
3. choose a = 1 and b = 3, so the new number is 2, and the whiteboard contains [2].
It's easy to see that after n - 1 operations, there will be left only one number. Your goal is to minimize it.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The only line of each test case contains one integer n (2 β€ n β€ 2 β
10^5) β the number of integers written on the board initially.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, in the first line, print the minimum possible number left on the board after n - 1 operations. Each of the next n - 1 lines should contain two integers β numbers a and b chosen and erased in each operation.
Example
Input
1
4
Output
2
2 4
3 3
3 1
Submitted Solution:
```
# cook your dish here
import heapq
import collections
from math import log2
import itertools
from functools import lru_cache
from sys import setrecursionlimit as srl
srl(2*10**6)
N = 200001
def solve(n):
print(2)
if n == 2:
return "{} {}".format(1,2)
print("{} {}".format(n,n-2))
print("{} {}".format(n-1,n-1))
i = n-3
while i >= 1 :
print("{} {}".format(i,i+2))
i-=1
t = int(input())
for tc in range(1,t+1):
n = int(input())
solve(n)
``` | instruction | 0 | 39,815 | 8 | 79,630 |
No | output | 1 | 39,815 | 8 | 79,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Numbers 1, 2, 3, ... n (each integer from 1 to n once) are written on a board. In one operation you can erase any two numbers a and b from the board and write one integer (a + b)/(2) rounded up instead.
You should perform the given operation n - 1 times and make the resulting number that will be left on the board as small as possible.
For example, if n = 4, the following course of action is optimal:
1. choose a = 4 and b = 2, so the new number is 3, and the whiteboard contains [1, 3, 3];
2. choose a = 3 and b = 3, so the new number is 3, and the whiteboard contains [1, 3];
3. choose a = 1 and b = 3, so the new number is 2, and the whiteboard contains [2].
It's easy to see that after n - 1 operations, there will be left only one number. Your goal is to minimize it.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The only line of each test case contains one integer n (2 β€ n β€ 2 β
10^5) β the number of integers written on the board initially.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, in the first line, print the minimum possible number left on the board after n - 1 operations. Each of the next n - 1 lines should contain two integers β numbers a and b chosen and erased in each operation.
Example
Input
1
4
Output
2
2 4
3 3
3 1
Submitted Solution:
```
import heapq
t=int(input())
for j in range(t):
n=int(input())
l=[i for i in range(1,n+1)]
heap=[]
heapq.heapify(heap)
for i in range(len(l)):
heapq.heappush(heap,-1*l[i])
for i in range(n-1):
a=heap[0]
b=heap[1]
heapq.heappop(heap)
heapq.heappop(heap)
heapq.heappush(heap,((a+b)//2))
print(-1*heap[0])
``` | instruction | 0 | 39,817 | 8 | 79,634 |
No | output | 1 | 39,817 | 8 | 79,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oscolcovo city has a campus consisting of n student dormitories, n universities and n military offices. Initially, the i-th dormitory belongs to the i-th university and is assigned to the i-th military office.
Life goes on and the campus is continuously going through some changes. The changes can be of four types:
1. University aj merges with university bj. After that all the dormitories that belonged to university bj are assigned to to university aj, and university bj disappears.
2. Military office cj merges with military office dj. After that all the dormitories that were assigned to military office dj, are assigned to military office cj, and military office dj disappears.
3. Students of university xj move in dormitories. Lets kxj is the number of dormitories that belong to this university at the time when the students move in. Then the number of students in each dormitory of university xj increases by kxj (note that the more dormitories belong to the university, the more students move in each dormitory of the university).
4. Military office number yj conducts raids on all the dormitories assigned to it and takes all students from there.
Thus, at each moment of time each dormitory is assigned to exactly one university and one military office. Initially, all the dormitory are empty.
Your task is to process the changes that take place in the campus and answer the queries, how many people currently live in dormitory qj.
Input
The first line contains two integers, n and m (1 β€ n, m β€ 5Β·105) β the number of dormitories and the number of queries, respectively.
Next m lines contain the queries, each of them is given in one of the following formats:
* Β«U aj bjΒ» β merging universities;
* Β«M cj djΒ» β merging military offices;
* Β«A xjΒ» β students of university xj moving in the dormitories;
* Β«Z yjΒ» β a raid in military office yj;
* Β«Q qjΒ» β a query asking the number of people in dormitory qj.
All the numbers in the queries are positive integers and do not exceed n. It is guaranteed that at the moment of the query the universities and military offices, that are present in the query, exist.
Output
In the i-th line print the answer to the i-th query asking the number of people in the dormitory.
Examples
Input
2 7
A 1
Q 1
U 1 2
A 1
Z 1
Q 1
Q 2
Output
1
0
2
Input
5 12
U 1 2
M 4 5
A 1
Q 1
A 3
A 4
Q 3
Q 4
Z 4
Q 4
A 5
Q 5
Output
2
1
1
0
1
Note
Consider the first sample test:
* In the first query university 1 owns only dormitory 1, so after the query dormitory 1 will have 1 student.
* After the third query university 1 owns dormitories 1 and 2.
* The fourth query increases by 2 the number of students living in dormitories 1 and 2 that belong to university number 1. After that 3 students live in the first dormitory and 2 students live in the second dormitory.
* At the fifth query the number of students living in dormitory 1, assigned to the military office 1, becomes zero.
Submitted Solution:
```
def campus(code, number):
for a in range(len(number)):
number[a] = int(number[a]) - 1
if len(number) == 1:
number = number[0]
else:
num1 = number[0]
num2 = number[1]
if code == 'A':
if len(uni_table[number]) == 1:
table[number] += 1
else:
for k in range(len(uni_table[number])):
for j in range(len(uni_table[number])):
table[k] += 1
elif code == 'U':
uni_table[num1].append(uni_table[num2])
elif code == 'M':
mil_table[num1].append(mil_table[num2])
elif code == 'Z':
table[number] = 0
else:
print(table[number])
# n = no. of dorms/unis/military offices
# m = no. of queries/changes
n, m = map(int,input().split())
table = [0]*n
uni_table = []
mil_table = []
for x in range(1,n+1):
uni_table.append([x])
mil_table.append([1])
# print(uni_table,'\n',mil_table,'\t',table)
for i in range(m):
query = list(map(str, input().split()))
# print(query)
campus(query[0],query[1:])
``` | instruction | 0 | 40,795 | 8 | 81,590 |
No | output | 1 | 40,795 | 8 | 81,591 |
Provide a correct Python 3 solution for this coding contest problem.
problem
The JOI building is a combination of regular hexagons with a side of 1 meter as shown in the figure. As Christmas is approaching, JOI decided to decorate the walls of the building with illuminations. However, since it is useless to illuminate the parts that cannot be seen from the outside, we decided to decorate only the walls that can be reached from the outside without passing through the building.
<image>
Figure: Example of JOI building layout
The figure above is an example of the layout of JOI's buildings as seen from above. The numbers in the regular hexagon represent the coordinates. A gray regular hexagon represents a place with a building, and a white regular hexagon represents a place without a building. In this example, the part indicated by the solid red line is the wall surface to be decorated with illumination, and the total length of the wall surface is 64 meters.
Given a map showing the layout of JOI's buildings, create a program to find the total length of the walls to decorate. However, the outside of the map can be freely moved, and it is not possible to pass between adjacent buildings.
input
Two integers W and H (1 β€ W β€ 100, 1 β€ H β€ 100) are written on the first line of the input file, separated by blanks. The following line H describes the layout of the JOI building. On the first line of i + (1 β€ i β€ H), W integers are written separated by blanks, and the jth (1 β€ j β€ W) integer is a regular hexagon with coordinates (j, i). It is 1 when there is a building in, and 0 when there is no building. Also, the input data given always has one or more buildings.
The map is described by the following rules.
* The westernmost regular hexagon in the northernmost row of the map is the coordinates (1, 1).
* The eastern adjacent regular hexagon adjacent to the regular hexagon with coordinates (x, y) is the coordinates (x + 1, y).
* When y is odd, the coordinates of the southwestern regular hexagon adjacent to the regular hexagon of coordinates (x, y) are (x, y + 1).
* When y is even, the coordinates of the southeastern regular hexagon adjacent to the regular hexagon of coordinates (x, y) are (x, y + 1).
output
Output the total length of the walls to be decorated with illuminations in one line.
Input / output example
Input example 1
8 4
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 0
1 0 1 0 1 1 1 1 1
0 1 1 0 1 0 1 0
Output example 1
64
Input / output example 1 corresponds to the example in the problem statement, and the total length of the wall surface decorated with illumination is 64 meters.
Input example 2
8 5
0 1 1 1 0 1 1 1
0 1 0 0 1 1 0 0
1 0 0 1 1 1 1 1 1
0 1 0 1 1 0 1 0
0 1 1 0 1 1 0 0
Output example 2
56
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
8 4
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 0
1 0 1 0 1 1 1 1
0 1 1 0 1 0 1 0
Output
64 | instruction | 0 | 41,134 | 8 | 82,268 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Illumination
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0569
"""
import sys
def solve(data, w, h):
invisible_area = set()
visible_area = set()
visited = set()
for y in range(h):
for x in range(w):
visited.add((x, y))
if data[y][x] == 1:
invisible_area.add((x, y))
is_visible = False
if (x, y) in invisible_area or (x, y) in visible_area:
continue
area = {(x, y)}
stack = [(x, y)]
while stack:
cx, cy = stack.pop()
visited.add((cx, cy))
dxy = ((cx, cy-1), (cx+1, cy-1), (cx-1, cy), (cx+1, cy), (cx, cy+1), (cx+1, cy+1)) if cy % 2 == 0 else \
((cx-1, cy-1), (cx, cy-1), (cx-1, cy), (cx+1, cy), (cx-1, cy+1), (cx, cy+1))
for nx, ny in dxy:
if not (0 <= nx < w) or not (0 <= ny < h):
is_visible = True
if 0 <= nx < w and 0 <= ny < h and data[ny][nx] == 0 and (nx, ny) not in visited:
stack.append((nx, ny))
area.add((nx, ny))
if is_visible:
visible_area |= area
else:
invisible_area |= area
ans = 0
for cy in range(h):
for cx in range(w):
if data[cy][cx] == 1:
dxy = ((cx, cy-1), (cx+1, cy-1), (cx-1, cy), (cx+1, cy), (cx, cy+1), (cx+1, cy+1)) if cy % 2 == 0 else \
((cx-1, cy-1), (cx, cy-1), (cx-1, cy), (cx+1, cy), (cx-1, cy+1), (cx, cy+1))
ans += 6 - sum((nx, ny) in invisible_area for nx, ny in dxy)
return ans
def main(args):
w, h = map(int, input().split())
data = [list(map(int, input().split())) for _ in range(h)]
ans = solve(data, w, h)
print(ans)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 41,134 | 8 | 82,269 |
Provide a correct Python 3 solution for this coding contest problem.
problem
The JOI building is a combination of regular hexagons with a side of 1 meter as shown in the figure. As Christmas is approaching, JOI decided to decorate the walls of the building with illuminations. However, since it is useless to illuminate the parts that cannot be seen from the outside, we decided to decorate only the walls that can be reached from the outside without passing through the building.
<image>
Figure: Example of JOI building layout
The figure above is an example of the layout of JOI's buildings as seen from above. The numbers in the regular hexagon represent the coordinates. A gray regular hexagon represents a place with a building, and a white regular hexagon represents a place without a building. In this example, the part indicated by the solid red line is the wall surface to be decorated with illumination, and the total length of the wall surface is 64 meters.
Given a map showing the layout of JOI's buildings, create a program to find the total length of the walls to decorate. However, the outside of the map can be freely moved, and it is not possible to pass between adjacent buildings.
input
Two integers W and H (1 β€ W β€ 100, 1 β€ H β€ 100) are written on the first line of the input file, separated by blanks. The following line H describes the layout of the JOI building. On the first line of i + (1 β€ i β€ H), W integers are written separated by blanks, and the jth (1 β€ j β€ W) integer is a regular hexagon with coordinates (j, i). It is 1 when there is a building in, and 0 when there is no building. Also, the input data given always has one or more buildings.
The map is described by the following rules.
* The westernmost regular hexagon in the northernmost row of the map is the coordinates (1, 1).
* The eastern adjacent regular hexagon adjacent to the regular hexagon with coordinates (x, y) is the coordinates (x + 1, y).
* When y is odd, the coordinates of the southwestern regular hexagon adjacent to the regular hexagon of coordinates (x, y) are (x, y + 1).
* When y is even, the coordinates of the southeastern regular hexagon adjacent to the regular hexagon of coordinates (x, y) are (x, y + 1).
output
Output the total length of the walls to be decorated with illuminations in one line.
Input / output example
Input example 1
8 4
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 0
1 0 1 0 1 1 1 1 1
0 1 1 0 1 0 1 0
Output example 1
64
Input / output example 1 corresponds to the example in the problem statement, and the total length of the wall surface decorated with illumination is 64 meters.
Input example 2
8 5
0 1 1 1 0 1 1 1
0 1 0 0 1 1 0 0
1 0 0 1 1 1 1 1 1
0 1 0 1 1 0 1 0
0 1 1 0 1 1 0 0
Output example 2
56
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
8 4
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 0
1 0 1 0 1 1 1 1
0 1 1 0 1 0 1 0
Output
64 | instruction | 0 | 41,135 | 8 | 82,270 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(1000000)
WALL = 100
w, h = map(int,input().split())
lst = [[WALL] + list(map(int,input().split())) + [WALL] for i in range(h)]
lst.insert(0, [WALL] * (w + 2))
lst.append([WALL] * (w + 2))
visited = [[0] * (w + 2) for _ in range(h + 2)]
hold = []
def search(x, y):
if lst[x][y] == WALL:
visited[x][y] = 3
return 3
if lst[x][y] == 1:
visited[x][y] = 2
return 2
visited[x][y] = 1
hold.append((x, y))
if not x % 2:
pairs = [(x - 1, y - 1), (x - 1, y), (x, y - 1), (x, y + 1), (x + 1, y - 1), (x + 1, y)]
else:
pairs = [(x - 1, y), (x - 1, y + 1), (x, y - 1), (x, y + 1), (x + 1, y), (x + 1, y + 1)]
ret = 0
for t in pairs:
tx, ty = t[0], t[1]
a = 0
if not visited[tx][ty]:
a = search(tx, ty)
elif visited[tx][ty] == 3:
a = 3
elif visited[tx][ty] == 2:
a = 2
if a > ret:
ret = a
return ret
for x in range(1, h + 1):
for y in range(1, w + 1):
if not visited[x][y] and not lst[x][y]:
stat = search(x, y)
for point in hold:
visited[point[0]][point[1]] = stat
hold.clear()
ans = 0
for x in range(1, h + 1):
for y in range(1, w + 1):
if lst[x][y]:
if not x % 2:
pairs = [(x - 1, y - 1), (x - 1, y), (x, y - 1), (x, y + 1), (x + 1, y - 1), (x + 1, y)]
else:
pairs = [(x - 1, y), (x - 1, y + 1), (x, y - 1), (x, y + 1), (x + 1, y), (x + 1, y + 1)]
for t in pairs:
tx, ty = t[0], t[1]
if (visited[tx][ty] in [0,3]) and (lst[tx][ty] in [WALL, 0]):
ans += 1
print(ans)
``` | output | 1 | 41,135 | 8 | 82,271 |
Provide a correct Python 3 solution for this coding contest problem.
problem
The JOI building is a combination of regular hexagons with a side of 1 meter as shown in the figure. As Christmas is approaching, JOI decided to decorate the walls of the building with illuminations. However, since it is useless to illuminate the parts that cannot be seen from the outside, we decided to decorate only the walls that can be reached from the outside without passing through the building.
<image>
Figure: Example of JOI building layout
The figure above is an example of the layout of JOI's buildings as seen from above. The numbers in the regular hexagon represent the coordinates. A gray regular hexagon represents a place with a building, and a white regular hexagon represents a place without a building. In this example, the part indicated by the solid red line is the wall surface to be decorated with illumination, and the total length of the wall surface is 64 meters.
Given a map showing the layout of JOI's buildings, create a program to find the total length of the walls to decorate. However, the outside of the map can be freely moved, and it is not possible to pass between adjacent buildings.
input
Two integers W and H (1 β€ W β€ 100, 1 β€ H β€ 100) are written on the first line of the input file, separated by blanks. The following line H describes the layout of the JOI building. On the first line of i + (1 β€ i β€ H), W integers are written separated by blanks, and the jth (1 β€ j β€ W) integer is a regular hexagon with coordinates (j, i). It is 1 when there is a building in, and 0 when there is no building. Also, the input data given always has one or more buildings.
The map is described by the following rules.
* The westernmost regular hexagon in the northernmost row of the map is the coordinates (1, 1).
* The eastern adjacent regular hexagon adjacent to the regular hexagon with coordinates (x, y) is the coordinates (x + 1, y).
* When y is odd, the coordinates of the southwestern regular hexagon adjacent to the regular hexagon of coordinates (x, y) are (x, y + 1).
* When y is even, the coordinates of the southeastern regular hexagon adjacent to the regular hexagon of coordinates (x, y) are (x, y + 1).
output
Output the total length of the walls to be decorated with illuminations in one line.
Input / output example
Input example 1
8 4
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 0
1 0 1 0 1 1 1 1 1
0 1 1 0 1 0 1 0
Output example 1
64
Input / output example 1 corresponds to the example in the problem statement, and the total length of the wall surface decorated with illumination is 64 meters.
Input example 2
8 5
0 1 1 1 0 1 1 1
0 1 0 0 1 1 0 0
1 0 0 1 1 1 1 1 1
0 1 0 1 1 0 1 0
0 1 1 0 1 1 0 0
Output example 2
56
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
8 4
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 0
1 0 1 0 1 1 1 1
0 1 1 0 1 0 1 0
Output
64 | instruction | 0 | 41,136 | 8 | 82,272 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(1000000)
WALL = 100
w, h = map(int,input().split())
lst = [[WALL] + list(map(int,input().split())) + [WALL] for i in range(h)]
lst.insert(0, [WALL] * (w + 2))
lst.append([WALL] * (w + 2))
visited = [[0] * (w + 2) for _ in range(h + 2)]
hold = []
app = hold.append
def search(x, y):
if lst[x][y] == WALL:
visited[x][y] = 3
return 3
if lst[x][y] == 1:
visited[x][y] = 2
return 2
visited[x][y] = 1
app((x, y))
if not x % 2:
pairs = [(x - 1, y - 1), (x - 1, y), (x, y - 1), (x, y + 1), (x + 1, y - 1), (x + 1, y)]
else:
pairs = [(x - 1, y), (x - 1, y + 1), (x, y - 1), (x, y + 1), (x + 1, y), (x + 1, y + 1)]
ret = 0
for t in pairs:
tx, ty = t[0], t[1]
v = visited[tx][ty]
a = 0
if not v:
a = search(tx, ty)
elif v == 3:
a = 3
elif v == 2:
a = 2
if a > ret:
ret = a
return ret
def main():
for x in range(1, h + 1):
for y in range(1, w + 1):
if not visited[x][y] and not lst[x][y]:
stat = search(x, y)
for point in hold:
visited[point[0]][point[1]] = stat
hold.clear()
ans = 0
for x in range(1, h + 1):
for y in range(1, w + 1):
if lst[x][y]:
if not x % 2:
pairs = [(x - 1, y - 1), (x - 1, y), (x, y - 1), (x, y + 1), (x + 1, y - 1), (x + 1, y)]
else:
pairs = [(x - 1, y), (x - 1, y + 1), (x, y - 1), (x, y + 1), (x + 1, y), (x + 1, y + 1)]
for t in pairs:
tx, ty = t[0], t[1]
if (visited[tx][ty] in [0,3]) and (lst[tx][ty] in [WALL, 0]):
ans += 1
print(ans)
main()
``` | output | 1 | 41,136 | 8 | 82,273 |
Provide a correct Python 3 solution for this coding contest problem.
problem
The JOI building is a combination of regular hexagons with a side of 1 meter as shown in the figure. As Christmas is approaching, JOI decided to decorate the walls of the building with illuminations. However, since it is useless to illuminate the parts that cannot be seen from the outside, we decided to decorate only the walls that can be reached from the outside without passing through the building.
<image>
Figure: Example of JOI building layout
The figure above is an example of the layout of JOI's buildings as seen from above. The numbers in the regular hexagon represent the coordinates. A gray regular hexagon represents a place with a building, and a white regular hexagon represents a place without a building. In this example, the part indicated by the solid red line is the wall surface to be decorated with illumination, and the total length of the wall surface is 64 meters.
Given a map showing the layout of JOI's buildings, create a program to find the total length of the walls to decorate. However, the outside of the map can be freely moved, and it is not possible to pass between adjacent buildings.
input
Two integers W and H (1 β€ W β€ 100, 1 β€ H β€ 100) are written on the first line of the input file, separated by blanks. The following line H describes the layout of the JOI building. On the first line of i + (1 β€ i β€ H), W integers are written separated by blanks, and the jth (1 β€ j β€ W) integer is a regular hexagon with coordinates (j, i). It is 1 when there is a building in, and 0 when there is no building. Also, the input data given always has one or more buildings.
The map is described by the following rules.
* The westernmost regular hexagon in the northernmost row of the map is the coordinates (1, 1).
* The eastern adjacent regular hexagon adjacent to the regular hexagon with coordinates (x, y) is the coordinates (x + 1, y).
* When y is odd, the coordinates of the southwestern regular hexagon adjacent to the regular hexagon of coordinates (x, y) are (x, y + 1).
* When y is even, the coordinates of the southeastern regular hexagon adjacent to the regular hexagon of coordinates (x, y) are (x, y + 1).
output
Output the total length of the walls to be decorated with illuminations in one line.
Input / output example
Input example 1
8 4
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 0
1 0 1 0 1 1 1 1 1
0 1 1 0 1 0 1 0
Output example 1
64
Input / output example 1 corresponds to the example in the problem statement, and the total length of the wall surface decorated with illumination is 64 meters.
Input example 2
8 5
0 1 1 1 0 1 1 1
0 1 0 0 1 1 0 0
1 0 0 1 1 1 1 1 1
0 1 0 1 1 0 1 0
0 1 1 0 1 1 0 0
Output example 2
56
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
8 4
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 0
1 0 1 0 1 1 1 1
0 1 1 0 1 0 1 0
Output
64 | instruction | 0 | 41,137 | 8 | 82,274 |
"Correct Solution:
```
def solve():
from collections import deque
from copy import deepcopy
W, H = map(int, input().split())
a = [[0]*(W+2)] + [list(map(int, ("0 "+input()+" 0").split())) for _ in [0]*H] + [[0]*(W+2)]
result = 0
visited = deepcopy(a)
dq = deque([(0, 0)])
append, popleft = dq.append, dq.popleft
while dq:
x, y = popleft()
neighbour = [(-1, 0), (1, 0)] + ([(-1, -1), (0, -1), (-1, 1), (0, 1)] if y%2 == 0 else [(0, -1), (1, -1), (0, 1), (1, 1)])
for dx, dy in neighbour:
nx, ny = dx+x, dy+y
if 0 <= nx < W+2 and 0 <= ny < H+2:
result += a[ny][nx]
for dx, dy in neighbour:
nx, ny = x+dx, y+dy
if 0 <= nx < W+2 and 0 <= ny < H+2 and visited[ny][nx] == 0:
visited[ny][nx] = 1
append((nx, ny))
print(result)
if __name__ == "__main__":
solve()
``` | output | 1 | 41,137 | 8 | 82,275 |
Provide a correct Python 3 solution for this coding contest problem.
problem
The JOI building is a combination of regular hexagons with a side of 1 meter as shown in the figure. As Christmas is approaching, JOI decided to decorate the walls of the building with illuminations. However, since it is useless to illuminate the parts that cannot be seen from the outside, we decided to decorate only the walls that can be reached from the outside without passing through the building.
<image>
Figure: Example of JOI building layout
The figure above is an example of the layout of JOI's buildings as seen from above. The numbers in the regular hexagon represent the coordinates. A gray regular hexagon represents a place with a building, and a white regular hexagon represents a place without a building. In this example, the part indicated by the solid red line is the wall surface to be decorated with illumination, and the total length of the wall surface is 64 meters.
Given a map showing the layout of JOI's buildings, create a program to find the total length of the walls to decorate. However, the outside of the map can be freely moved, and it is not possible to pass between adjacent buildings.
input
Two integers W and H (1 β€ W β€ 100, 1 β€ H β€ 100) are written on the first line of the input file, separated by blanks. The following line H describes the layout of the JOI building. On the first line of i + (1 β€ i β€ H), W integers are written separated by blanks, and the jth (1 β€ j β€ W) integer is a regular hexagon with coordinates (j, i). It is 1 when there is a building in, and 0 when there is no building. Also, the input data given always has one or more buildings.
The map is described by the following rules.
* The westernmost regular hexagon in the northernmost row of the map is the coordinates (1, 1).
* The eastern adjacent regular hexagon adjacent to the regular hexagon with coordinates (x, y) is the coordinates (x + 1, y).
* When y is odd, the coordinates of the southwestern regular hexagon adjacent to the regular hexagon of coordinates (x, y) are (x, y + 1).
* When y is even, the coordinates of the southeastern regular hexagon adjacent to the regular hexagon of coordinates (x, y) are (x, y + 1).
output
Output the total length of the walls to be decorated with illuminations in one line.
Input / output example
Input example 1
8 4
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 0
1 0 1 0 1 1 1 1 1
0 1 1 0 1 0 1 0
Output example 1
64
Input / output example 1 corresponds to the example in the problem statement, and the total length of the wall surface decorated with illumination is 64 meters.
Input example 2
8 5
0 1 1 1 0 1 1 1
0 1 0 0 1 1 0 0
1 0 0 1 1 1 1 1 1
0 1 0 1 1 0 1 0
0 1 1 0 1 1 0 0
Output example 2
56
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
8 4
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 0
1 0 1 0 1 1 1 1
0 1 1 0 1 0 1 0
Output
64 | instruction | 0 | 41,138 | 8 | 82,276 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(100000)
W, H = map(int, input().split())
m = [list(map(int, input().split())) for i in range(H)]
dx = [[1, 1, 1, 0, -1, 0], [0, 1, 0, -1, -1, -1]]
dy = [-1, 0, 1, 1, 0, -1]
def dfs(x, y):
if m[y][x] != 0:
return
m[y][x] = 2
for xx, yy in zip(dx[y % 2], dy):
tx, ty = x + xx, y + yy
if 0 <= tx < W and 0 <= ty < H:
dfs(tx, ty)
for x in range(W):
dfs(x, 0)
dfs(x, H - 1)
for y in range(H):
dfs(0, y)
dfs(W - 1, y)
from itertools import product
n = 0
for x, y in product(range(W), range(H)):
if m[y][x] != 1:
continue
fn = n
for xx, yy in zip(dx[y % 2], dy):
tx, ty = x + xx, y + yy
if 0 <= tx < W and 0 <= ty < H:
if m[ty][tx] == 2:
n += 1
else:
n += 1
print(n)
``` | output | 1 | 41,138 | 8 | 82,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[Sakuzyo - Imprinting](https://www.youtube.com/watch?v=55Ca6av1kAY)
A.R.C. Markland-N is a tall building with n floors numbered from 1 to n. Between each two adjacent floors in the building, there is a staircase connecting them.
It's lunchtime for our sensei Colin "ConneR" Neumann Jr, and he's planning for a location to enjoy his meal.
ConneR's office is at floor s of the building. On each floor (including floor s, of course), there is a restaurant offering meals. However, due to renovations being in progress, k of the restaurants are currently closed, and as a result, ConneR can't enjoy his lunch there.
CooneR wants to reach a restaurant as quickly as possible to save time. What is the minimum number of staircases he needs to walk to reach a closest currently open restaurant.
Please answer him quickly, and you might earn his praise and even enjoy the lunch with him in the elegant Neumanns' way!
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases in the test. Then the descriptions of t test cases follow.
The first line of a test case contains three integers n, s and k (2 β€ n β€ 10^9, 1 β€ s β€ n, 1 β€ k β€ min(n-1, 1000)) β respectively the number of floors of A.R.C. Markland-N, the floor where ConneR is in, and the number of closed restaurants.
The second line of a test case contains k distinct integers a_1, a_2, β¦, a_k (1 β€ a_i β€ n) β the floor numbers of the currently closed restaurants.
It is guaranteed that the sum of k over all test cases does not exceed 1000.
Output
For each test case print a single integer β the minimum number of staircases required for ConneR to walk from the floor s to a floor with an open restaurant.
Example
Input
5
5 2 3
1 2 3
4 3 3
4 1 2
10 2 6
1 2 3 4 5 7
2 1 1
2
100 76 8
76 75 36 67 41 74 10 77
Output
2
0
4
0
2
Note
In the first example test case, the nearest floor with an open restaurant would be the floor 4.
In the second example test case, the floor with ConneR's office still has an open restaurant, so Sensei won't have to go anywhere.
In the third example test case, the closest open restaurant is on the 6-th floor. | instruction | 0 | 41,365 | 8 | 82,730 |
Tags: binary search, brute force, implementation
Correct Solution:
```
t = int(input())
for j in range(t):
b = input()
b = list(map(int, b.split()))
n = b[0]
s = b[1]
g = b[2]
a = input()
k = list(map(int, a.split()))
d = {}
for i in k:
d[i] = 1
if s not in k:
print(0)
else:
step = 1
while True:
if s-step >0 and s-step not in d.keys():
print(step)
break
if s+step <= n and s+step not in d.keys():
print(step)
break
step += 1
``` | output | 1 | 41,365 | 8 | 82,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[Sakuzyo - Imprinting](https://www.youtube.com/watch?v=55Ca6av1kAY)
A.R.C. Markland-N is a tall building with n floors numbered from 1 to n. Between each two adjacent floors in the building, there is a staircase connecting them.
It's lunchtime for our sensei Colin "ConneR" Neumann Jr, and he's planning for a location to enjoy his meal.
ConneR's office is at floor s of the building. On each floor (including floor s, of course), there is a restaurant offering meals. However, due to renovations being in progress, k of the restaurants are currently closed, and as a result, ConneR can't enjoy his lunch there.
CooneR wants to reach a restaurant as quickly as possible to save time. What is the minimum number of staircases he needs to walk to reach a closest currently open restaurant.
Please answer him quickly, and you might earn his praise and even enjoy the lunch with him in the elegant Neumanns' way!
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases in the test. Then the descriptions of t test cases follow.
The first line of a test case contains three integers n, s and k (2 β€ n β€ 10^9, 1 β€ s β€ n, 1 β€ k β€ min(n-1, 1000)) β respectively the number of floors of A.R.C. Markland-N, the floor where ConneR is in, and the number of closed restaurants.
The second line of a test case contains k distinct integers a_1, a_2, β¦, a_k (1 β€ a_i β€ n) β the floor numbers of the currently closed restaurants.
It is guaranteed that the sum of k over all test cases does not exceed 1000.
Output
For each test case print a single integer β the minimum number of staircases required for ConneR to walk from the floor s to a floor with an open restaurant.
Example
Input
5
5 2 3
1 2 3
4 3 3
4 1 2
10 2 6
1 2 3 4 5 7
2 1 1
2
100 76 8
76 75 36 67 41 74 10 77
Output
2
0
4
0
2
Note
In the first example test case, the nearest floor with an open restaurant would be the floor 4.
In the second example test case, the floor with ConneR's office still has an open restaurant, so Sensei won't have to go anywhere.
In the third example test case, the closest open restaurant is on the 6-th floor. | instruction | 0 | 41,366 | 8 | 82,732 |
Tags: binary search, brute force, implementation
Correct Solution:
```
for _ in range(int(input())):
n,s,k = map(int,input().split())
l = list(map(int,input().split()))
h = {i:1 for i in l}
g = []
for i in range(s,n+1):
if i not in h:
g.append(i-s)
break
for j in range(s,0,-1):
if j not in h:
g.append(s-j)
break
print(min(g))
``` | output | 1 | 41,366 | 8 | 82,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[Sakuzyo - Imprinting](https://www.youtube.com/watch?v=55Ca6av1kAY)
A.R.C. Markland-N is a tall building with n floors numbered from 1 to n. Between each two adjacent floors in the building, there is a staircase connecting them.
It's lunchtime for our sensei Colin "ConneR" Neumann Jr, and he's planning for a location to enjoy his meal.
ConneR's office is at floor s of the building. On each floor (including floor s, of course), there is a restaurant offering meals. However, due to renovations being in progress, k of the restaurants are currently closed, and as a result, ConneR can't enjoy his lunch there.
CooneR wants to reach a restaurant as quickly as possible to save time. What is the minimum number of staircases he needs to walk to reach a closest currently open restaurant.
Please answer him quickly, and you might earn his praise and even enjoy the lunch with him in the elegant Neumanns' way!
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases in the test. Then the descriptions of t test cases follow.
The first line of a test case contains three integers n, s and k (2 β€ n β€ 10^9, 1 β€ s β€ n, 1 β€ k β€ min(n-1, 1000)) β respectively the number of floors of A.R.C. Markland-N, the floor where ConneR is in, and the number of closed restaurants.
The second line of a test case contains k distinct integers a_1, a_2, β¦, a_k (1 β€ a_i β€ n) β the floor numbers of the currently closed restaurants.
It is guaranteed that the sum of k over all test cases does not exceed 1000.
Output
For each test case print a single integer β the minimum number of staircases required for ConneR to walk from the floor s to a floor with an open restaurant.
Example
Input
5
5 2 3
1 2 3
4 3 3
4 1 2
10 2 6
1 2 3 4 5 7
2 1 1
2
100 76 8
76 75 36 67 41 74 10 77
Output
2
0
4
0
2
Note
In the first example test case, the nearest floor with an open restaurant would be the floor 4.
In the second example test case, the floor with ConneR's office still has an open restaurant, so Sensei won't have to go anywhere.
In the third example test case, the closest open restaurant is on the 6-th floor. | instruction | 0 | 41,367 | 8 | 82,734 |
Tags: binary search, brute force, implementation
Correct Solution:
```
t=int(input())
for _ in range(0,t):
n,s,k=map(int,input().split())
a=list(map(int,input().split()))
if s not in a:
print(0)
else:
p=0
d=0
for i in range(s,n+1):
if i not in a:
p=i-s
break
for i in range(s-1,0,-1):
if i not in a:
d=s-i
break
if p==0:
print(d)
elif d==0:
print(p)
else:
print(min(p,d))
``` | output | 1 | 41,367 | 8 | 82,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[Sakuzyo - Imprinting](https://www.youtube.com/watch?v=55Ca6av1kAY)
A.R.C. Markland-N is a tall building with n floors numbered from 1 to n. Between each two adjacent floors in the building, there is a staircase connecting them.
It's lunchtime for our sensei Colin "ConneR" Neumann Jr, and he's planning for a location to enjoy his meal.
ConneR's office is at floor s of the building. On each floor (including floor s, of course), there is a restaurant offering meals. However, due to renovations being in progress, k of the restaurants are currently closed, and as a result, ConneR can't enjoy his lunch there.
CooneR wants to reach a restaurant as quickly as possible to save time. What is the minimum number of staircases he needs to walk to reach a closest currently open restaurant.
Please answer him quickly, and you might earn his praise and even enjoy the lunch with him in the elegant Neumanns' way!
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases in the test. Then the descriptions of t test cases follow.
The first line of a test case contains three integers n, s and k (2 β€ n β€ 10^9, 1 β€ s β€ n, 1 β€ k β€ min(n-1, 1000)) β respectively the number of floors of A.R.C. Markland-N, the floor where ConneR is in, and the number of closed restaurants.
The second line of a test case contains k distinct integers a_1, a_2, β¦, a_k (1 β€ a_i β€ n) β the floor numbers of the currently closed restaurants.
It is guaranteed that the sum of k over all test cases does not exceed 1000.
Output
For each test case print a single integer β the minimum number of staircases required for ConneR to walk from the floor s to a floor with an open restaurant.
Example
Input
5
5 2 3
1 2 3
4 3 3
4 1 2
10 2 6
1 2 3 4 5 7
2 1 1
2
100 76 8
76 75 36 67 41 74 10 77
Output
2
0
4
0
2
Note
In the first example test case, the nearest floor with an open restaurant would be the floor 4.
In the second example test case, the floor with ConneR's office still has an open restaurant, so Sensei won't have to go anywhere.
In the third example test case, the closest open restaurant is on the 6-th floor. | instruction | 0 | 41,368 | 8 | 82,736 |
Tags: binary search, brute force, implementation
Correct Solution:
```
for _ in range(int(input())):
n,s,k=map(int,input().split())
a = list(map(int,input().split()))
m1,m2=1001,1001
if s not in a:
print(0)
continue
for i in range(s-1,0,-1):
if i not in a:
m1=s-i
break
for i in range(s+1,n+1):
if i not in a:
m2=i-s
break
print(min(m1,m2))
``` | output | 1 | 41,368 | 8 | 82,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[Sakuzyo - Imprinting](https://www.youtube.com/watch?v=55Ca6av1kAY)
A.R.C. Markland-N is a tall building with n floors numbered from 1 to n. Between each two adjacent floors in the building, there is a staircase connecting them.
It's lunchtime for our sensei Colin "ConneR" Neumann Jr, and he's planning for a location to enjoy his meal.
ConneR's office is at floor s of the building. On each floor (including floor s, of course), there is a restaurant offering meals. However, due to renovations being in progress, k of the restaurants are currently closed, and as a result, ConneR can't enjoy his lunch there.
CooneR wants to reach a restaurant as quickly as possible to save time. What is the minimum number of staircases he needs to walk to reach a closest currently open restaurant.
Please answer him quickly, and you might earn his praise and even enjoy the lunch with him in the elegant Neumanns' way!
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases in the test. Then the descriptions of t test cases follow.
The first line of a test case contains three integers n, s and k (2 β€ n β€ 10^9, 1 β€ s β€ n, 1 β€ k β€ min(n-1, 1000)) β respectively the number of floors of A.R.C. Markland-N, the floor where ConneR is in, and the number of closed restaurants.
The second line of a test case contains k distinct integers a_1, a_2, β¦, a_k (1 β€ a_i β€ n) β the floor numbers of the currently closed restaurants.
It is guaranteed that the sum of k over all test cases does not exceed 1000.
Output
For each test case print a single integer β the minimum number of staircases required for ConneR to walk from the floor s to a floor with an open restaurant.
Example
Input
5
5 2 3
1 2 3
4 3 3
4 1 2
10 2 6
1 2 3 4 5 7
2 1 1
2
100 76 8
76 75 36 67 41 74 10 77
Output
2
0
4
0
2
Note
In the first example test case, the nearest floor with an open restaurant would be the floor 4.
In the second example test case, the floor with ConneR's office still has an open restaurant, so Sensei won't have to go anywhere.
In the third example test case, the closest open restaurant is on the 6-th floor. | instruction | 0 | 41,369 | 8 | 82,738 |
Tags: binary search, brute force, implementation
Correct Solution:
```
for _ in range(int(input())):
n,s,k = map(int, input().split())
se = set (map(int, input().split()))
m1 = 99999999999999999
m2 = 99999999999999999
for i in range(s, 0, -1):
if i not in se:
m1 = s-i
break
for i in range(s, n+1):
if i not in se:
m2 = i-s
break
print(min(m1, m2))
``` | output | 1 | 41,369 | 8 | 82,739 |
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