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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10). | instruction | 0 | 7,636 | 8 | 15,272 |
Tags: implementation
Correct Solution:
```
a,b,c=input().split()
a,b,c=int(a),int(b),int(c)
v=[]
s=0
d={}
min=50
max=0
for i in range(c):
v.append(input().split())
for j in range(int(v[i][0]),int(v[i][1])+1):
if int(v[i][0])<min:
min=int(v[i][0])
if int(v[i][1])>max:
max=int(v[i][1])
if j not in d or int(d[j])>int(v[i][2]):
d[j]=int(v[i][2])
for i in range(1,min):
d[i]=b
for i in range(max+1,a+1):
d[i]=b
for i in range(1,a+1):
s+=int(d[i])*int(d[i])
print(s)
``` | output | 1 | 7,636 | 8 | 15,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10). | instruction | 0 | 7,637 | 8 | 15,274 |
Tags: implementation
Correct Solution:
```
n, h, m = map(int, input().split())
a = [list(map(int, input().split())) for i in range(m)]
a.sort(key=lambda x: x[2])
count = 0
mest = 0
b = [0 for i in range(n)]
for i in range(len(a)):
for j in range(a[i][0] - 1, a[i][1]):
if b[j] == 0:
b[j] = 1
count += (min(h, a[i][2]) ** 2)
mest += 1
print(count + (n - mest) * (h ** 2))
``` | output | 1 | 7,637 | 8 | 15,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10).
Submitted Solution:
```
n, h, m = [int(i) for i in input().split()]
t = [h for _ in range(n)]
for _ in range(m):
l, r, x = [int(j) for j in input().split()]
for a in range(l - 1, r):
t[a] = min(t[a], x)
print(sum(map(lambda w: w * w, t)))
``` | instruction | 0 | 7,638 | 8 | 15,276 |
Yes | output | 1 | 7,638 | 8 | 15,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10).
Submitted Solution:
```
n, h, m = [int(i) for i in input().split()]
a = [h] * n
for _ in range(m):
l, r, x = [int(i) for i in input().split()]
for j in range(l - 1, r):
a[j] = min(a[j], x)
print(sum([i * i for i in a]))
``` | instruction | 0 | 7,639 | 8 | 15,278 |
Yes | output | 1 | 7,639 | 8 | 15,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10).
Submitted Solution:
```
n,h,m = map(int,input().split())
a = {}
for i in range(n):
a[i+1] = h
b = {}
for i in range(m):
l,r,x = map(int,input().split())
i = l
while i < r + 1:
b[i] = x
if b[i] < a[i]:
a[i] = b[i]
i = i + 1
b = []
for i in a.values():
b.append(i)
s = 0
for i in range(len(b)):
s = s + b[i] ** 2
print(s)
``` | instruction | 0 | 7,640 | 8 | 15,280 |
Yes | output | 1 | 7,640 | 8 | 15,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10).
Submitted Solution:
```
n, h, m=map(int, input().split())
a=[3000]*55
for _ in range(m):
l, r, x=map(int, input().split())
for i in range(l, r+1):
if x*x<a[i]:
a[i]=x*x
s=0
for i in range(1, n+1):
if a[i]==3000:
s+=h*h
else:
s+=a[i]
print(s)
``` | instruction | 0 | 7,641 | 8 | 15,282 |
Yes | output | 1 | 7,641 | 8 | 15,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10).
Submitted Solution:
```
n, h, m=map(int, input().split())
s=0
a=[]
for i in range(n):
a+=[[0,0]]
for i in range(m):
l, r, x=map(int, input().split())
for j in range(l-1,r):
if a[j][1]==0:
a[j][0]=1;a[j][1]=x
else:
if x<=a[j][1]:
a[j][0]=1;a[j][1]=x
for i in range(n):
if a[i][0]==0:
a[i][1]=h
s+=(a[i][1]*a[i][1])
print(s)
``` | instruction | 0 | 7,642 | 8 | 15,284 |
No | output | 1 | 7,642 | 8 | 15,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10).
Submitted Solution:
```
n, h, m = map(int, input().split())
a = [h] * n
for i in range(m):
l, r, x = map(int, input().split())
for i in range(l, r + 1):
a[i- 1] = min(a[l - 1], x)
print(sum(map(lambda x: x ** 2, a)))
``` | instruction | 0 | 7,643 | 8 | 15,286 |
No | output | 1 | 7,643 | 8 | 15,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10).
Submitted Solution:
```
from sys import stdin
def solve():
#stdin = open("G. Zoning Restrictions.txt")
N, H, M = map(int, stdin.readline().split())
fine = [[0,0] for _ in range(N+1)]
for _ in range(M):
l, r, x, c = map(int, stdin.readline().split())
for i in range(l, r+1):
if not fine[i][0]:
fine[i][0] = c
fine[i][1] = x
continue
fine[i][0] = max(fine[i][0], c)
fine[i][1] = max(fine[i][1], x)
ans = 0
for i in range(1,N+1):
if fine[i][0] == 0 or fine[i][1] > H:
ans += H*H
continue
ans += max(H*H - fine[i][0], fine[i][1]**2)
print (ans)
if __name__ == "__main__":
solve()
``` | instruction | 0 | 7,644 | 8 | 15,288 |
No | output | 1 | 7,644 | 8 | 15,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.
In each spot, if a house has height a, you will gain a^2 dollars from it.
The city has m zoning restrictions. The i-th restriction says that the tallest house from spots l_i to r_i (inclusive) must be at most x_i.
You would like to build houses to maximize your profit. Determine the maximum profit possible.
Input
The first line contains three integers n, h, and m (1 β€ n,h,m β€ 50) β the number of spots, the maximum height, and the number of restrictions.
Each of the next m lines contains three integers l_i, r_i, and x_i (1 β€ l_i β€ r_i β€ n, 0 β€ x_i β€ h) β left and right limits (inclusive) of the i-th restriction and the maximum possible height in that range.
Output
Print a single integer, the maximum profit you can make.
Examples
Input
3 3 3
1 1 1
2 2 3
3 3 2
Output
14
Input
4 10 2
2 3 8
3 4 7
Output
262
Note
In the first example, there are 3 houses, the maximum height of a house is 3, and there are 3 restrictions. The first restriction says the tallest house between 1 and 1 must be at most 1. The second restriction says the tallest house between 2 and 2 must be at most 3. The third restriction says the tallest house between 3 and 3 must be at most 2.
In this case, it is optimal to build houses with heights [1, 3, 2]. This fits within all the restrictions. The total profit in this case is 1^2 + 3^2 + 2^2 = 14.
In the second example, there are 4 houses, the maximum height of a house is 10, and there are 2 restrictions. The first restriction says the tallest house from 2 to 3 must be at most 8. The second restriction says the tallest house from 3 to 4 must be at most 7.
In this case, it's optimal to build houses with heights [10, 8, 7, 7]. We get a profit of 10^2+8^2+7^2+7^2 = 262. Note that there are two restrictions on house 3 and both of them must be satisfied. Also, note that even though there isn't any explicit restrictions on house 1, we must still limit its height to be at most 10 (h=10).
Submitted Solution:
```
n,h,m = map(int,input().split())
sum ,pre,cost= 0,0,0
count = 0
for i in range(m):
a,b,c = map(int,input().split())
if a!=pre:
count += (b-a+1)
sum+=(b-a+1)*c**2
pre = b
cost = c
else:
count+=(b-a)
sum+=(b-a)*c**2-cost**2+min(cost,c)**2
pre = b
cost = c
print((n-count)*h**2+sum)
``` | instruction | 0 | 7,645 | 8 | 15,290 |
No | output | 1 | 7,645 | 8 | 15,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010.
Submitted Solution:
```
kol = int(input())
sp = list(input())
a = ['0'] * 10
for i in sp:
if i == 'L':
a[a.index('0')] = '1'
elif i == 'R':
for j in range(9, -1, -1):
if a[j] == '0':
a[j] = '1'
break
else:
a[int(i)] = '0'
print(''.join(a))
``` | instruction | 0 | 7,658 | 8 | 15,316 |
Yes | output | 1 | 7,658 | 8 | 15,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010.
Submitted Solution:
```
current = [0 for _ in range(10)]
num_events = input()
events = input()
for i in range(int(num_events)):
if events[i] == 'L':
for j in range(len(current)):
if current[j] == 0:
current[j] = 1
break
elif events[i] == 'R':
for j in reversed(range(len(current))):
if current[j] == 0:
current[j] = 1
break
else:
current[int(events[i])] = 0
def convert(list):
res = "".join(map(str, list))
return res
print(convert(current))
``` | instruction | 0 | 7,659 | 8 | 15,318 |
Yes | output | 1 | 7,659 | 8 | 15,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010.
Submitted Solution:
```
n = int(input())
string = input()
rooms_status = [0 for i in range(10)]
for char in string:
if char == 'L':
for i in range(10):
if rooms_status[i] == 0:
rooms_status[i] = 1
break
elif char == 'R':
for i in range(9, -1, -1):
if rooms_status[i] == 0:
rooms_status[i] = 1
break
else:
rooms_status[int(char)] = 0
for room_status in rooms_status:
print(room_status, end='')
print()
``` | instruction | 0 | 7,660 | 8 | 15,320 |
Yes | output | 1 | 7,660 | 8 | 15,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010.
Submitted Solution:
```
n = int(input())
room = [0 for i in range(10)]
s = input()
for i in range(n):
if s[i] == "L":
for j in range(10):
if room[j] == 0:
room[j] = 1
break
elif s[i] == "R":
for j in range(9, -1, -1):
if room[j] == 0:
room[j] = 1
break
else:
room[int(s[i])] = 0
print("".join(map(str, room)))
``` | instruction | 0 | 7,661 | 8 | 15,322 |
Yes | output | 1 | 7,661 | 8 | 15,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010.
Submitted Solution:
```
x=int(input())
k=list(input())
p=['0' for i in range(10)]
for i in range(x):
if k[i]=='R':
for i in range(9,-1,-1):
if p[i]=='0':
p[i]='1'
print(p)
break
elif k[i]=='L':
for i in range(10):
if p[i]=='0':
p[i]='1'
print(p)
break
else:p[int(k[i])]='0'
for i in p:
print(i,end='')
``` | instruction | 0 | 7,662 | 8 | 15,324 |
No | output | 1 | 7,662 | 8 | 15,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010.
Submitted Solution:
```
current = [0 for _ in range(10)]
num_events = input()
events = input()
for i in range(int(num_events)):
if events[i] == 'L':
for j in range(len(current)):
if current[j] == 0:
current[j] = 1
break
elif events[i] == 'R':
for j in reversed(range(len(current))):
if current[j] == 0:
current[j] = 1
break
else:
current[int(events[i])] = 0
print(current)
def convert(list):
res = int("".join(map(str, list)))
return res
print(convert(current))
``` | instruction | 0 | 7,663 | 8 | 15,326 |
No | output | 1 | 7,663 | 8 | 15,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010.
Submitted Solution:
```
n = int(input())
a = [0]*(10)
s= input()
def get_r_index():
ind = -1
while a[ind]!=0:
ind -= 1
return ind
def get_l_index():
ind = 0
while a[ind]!=0:
ind += 1
return ind
for i in range(n):
if s[i]=='R':
a[get_r_index()] = 1
elif s[i]=='L':
a[get_l_index()] = 1
else:
pos = int(s[i])
a[pos] = 0
for i in a:
print(i, end=' ')
``` | instruction | 0 | 7,664 | 8 | 15,328 |
No | output | 1 | 7,664 | 8 | 15,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010.
Submitted Solution:
```
n = int(input())
m = input()
k = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
for i in range(0, n):
if m[i] == "L":
k[k.index(0)] = 1
elif m[i] == "R":
k.reverse()
k[k.index(0)] = 1
k.reverse()
else:
k[int(m[i])] = 0
print(k)
``` | instruction | 0 | 7,665 | 8 | 15,330 |
No | output | 1 | 7,665 | 8 | 15,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0 | instruction | 0 | 7,751 | 8 | 15,502 |
Tags: greedy, implementation, sortings
Correct Solution:
```
for _ in range(int(input())):
n,k = input().split()
lis=list(map(int, input().split()))
lis.sort(reverse=True)
for i in range(1,int(k)+1):
if(i<int(n)):
lis[0]+=lis[i]
print(lis[0])
``` | output | 1 | 7,751 | 8 | 15,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0 | instruction | 0 | 7,752 | 8 | 15,504 |
Tags: greedy, implementation, sortings
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key,lru_cache
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# import sys
# input = sys.stdin.readline
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip().split()]
def st():return str(input().rstrip())[2:-1]
def val():return int(input().rstrip())
def li2():return [str(i)[2:-1] for i in input().rstrip().split()]
def li3():return [int(i) for i in st()]
for _ in range(val()):
n, k = li()
l = sorted(li())
for i in range(k):
l[-1] += l[n - i - 2]
print(l[-1])
``` | output | 1 | 7,752 | 8 | 15,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0 | instruction | 0 | 7,753 | 8 | 15,506 |
Tags: greedy, implementation, sortings
Correct Solution:
```
import sys
import heapq
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
N, K = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
A.sort(reverse=True)
ans = 0
for i in range(min(N, K + 1)):
ans += A[i]
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 7,753 | 8 | 15,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0 | instruction | 0 | 7,754 | 8 | 15,508 |
Tags: greedy, implementation, sortings
Correct Solution:
```
from sys import stdin
def barrels(k, a):
score, *a = sorted(a)[::-1]
for i in range(k):
if i > len(a) - 1:
break
score += a[i]
print(score)
if __name__ == "__main__":
t = int(stdin.readline())
cases = []
for _ in range(t):
n, k = (int(s) for s in stdin.readline().split())
a = [int(s) for s in stdin.readline().split()]
cases.append((k, a))
for c in cases:
barrels(*c)
``` | output | 1 | 7,754 | 8 | 15,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0 | instruction | 0 | 7,755 | 8 | 15,510 |
Tags: greedy, implementation, sortings
Correct Solution:
```
t = int(input())
for i in range(t):
n, k = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
s = 0
for j in range(n - k - 1, n):
s += a[j]
print(s)
``` | output | 1 | 7,755 | 8 | 15,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0 | instruction | 0 | 7,756 | 8 | 15,512 |
Tags: greedy, implementation, sortings
Correct Solution:
```
from sys import stdin, stdout
import math,sys,heapq
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import random
import bisect as bi
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')):
#------------------Sublime--------------------------------------#
sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def I():return (int(input()))
def In():return(map(int,input().split()))
else:
#------------------PYPY FAst I/o--------------------------------#
def I():return (int(stdin.readline()))
def In():return(map(int,stdin.readline().split()))
#sys.setrecursionlimit(1500)
def dict(a):
d={}
for x in a:
if d.get(x,-1)!=-1:
d[x]+=1
else:
d[x]=1
return d
def find_gt(a, x):
'Find leftmost value greater than x'
i = bi.bisect_left(a, x)
if i != len(a):
return i
else:
return -1
def main():
try:
n,k=In()
l=list(In())
cnt=0
su=0
l.sort(reverse=True)
temp=[]
for x in range(1,n):
if l[x]!=0:
if k:
su+=l[x]
k-=1
else:
break
print(l[0]+su)
except:
pass
M = 998244353
P = 1000000007
if __name__ == '__main__':
for _ in range(I()):main()
#for _ in range(1):main()
``` | output | 1 | 7,756 | 8 | 15,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0 | instruction | 0 | 7,757 | 8 | 15,514 |
Tags: greedy, implementation, sortings
Correct Solution:
```
from math import gcd
from sys import stdout,stdin
for _ in range(int(stdin.readline())):
n,k = map(int,input().split())
ls = [int(x) for x in input().split()]
ls = sorted(ls,reverse=True)
s =0
pref =[]
for i in ls:
s+=i
pref.append(s)
#print(pref)
print(pref[k])
``` | output | 1 | 7,757 | 8 | 15,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0 | instruction | 0 | 7,758 | 8 | 15,516 |
Tags: greedy, implementation, sortings
Correct Solution:
```
from math import *
t=int(input())
while t:
t=t-1
n,k=map(int,input().split())
#n=int(input())
a=list(map(int,input().split()))
#
#s=input()
a.sort()
j=n-2
for i in range(k):
a[n-1]+=a[j]
j=j-1
print(a[n-1])
``` | output | 1 | 7,758 | 8 | 15,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
Submitted Solution:
```
import sys
infile = sys.stdin
next(infile)
for line in infile:
l1 = list(map(int,line.split()))
n = l1[0]
k = l1[1]
l2 = list(map(int,infile.readline().split()))
l2.sort()
if l2[-1] == 0:
print(0)
else:
ans = 0
n = min(len(l2),k+1)
for i in range(n):
ans += l2[len(l2)-1-i]
print(ans)
``` | instruction | 0 | 7,759 | 8 | 15,518 |
Yes | output | 1 | 7,759 | 8 | 15,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
Submitted Solution:
```
def func2( number ):
if number == 1:
return ( -1, 0, 0 )
if number == 2:
return ( -1, 0, 0 )
if number == 3:
return ( 0, 0, 1 )
if number == 4:
return ( -1, 0, 0 )
if number == 5:
return ( 0, 1, 0 )
if number == 6:
return ( 0, 0, 2 )
if number == 7:
return ( 1, 0, 0 )
if number == 8:
return ( 0, 1, 1 )
if number == 9:
return ( 0, 0, 3 )
if number == 10:
return ( 0, 2, 0 )
if number == 11:
return ( 0, 1, 2 )
if number == 12:
return ( 0, 0, 4 )
if number == 13:
return ( 0, 2, 1 )
if number == 14:
return ( 2, 0, 0 )
if number == 15:
return ( 0, 3, 0 )
if number == 16:
return ( 0, 2, 2 )
if number == 17:
return ( 1, 2, 0 )
if number == 18:
return ( 0, 3, 1 )
if number == 19:
return ( 0, 3, 3 )
if number == 20:
return ( 0, 4, 0 )
def func():
n = int( input() )
n7 = 0
n5 = 0
n3 = 0
if ( n <= 10 ):
(n7, n5, n3 ) = func2( n )
else:
x = n % 10
y = n // 10
if x <= 4:
y = y - 1
x = x + 10
(n7, n5, n3 ) = func2( x )
n5 = n5 + ( y * 2 )
if ( n7 == -1 ):
print ( -1 )
return
print ( n3, n5, n7 )
return
def codeforces1430ProblemB():
x = [ int(i) for i in input().split(" ") ]
n = x[0]
k = x[1]
tanks = []
tanks = [ int(i) for i in input().split(" ") ]
tanks = sorted( tanks )
maxDiff = tanks[n-1]
# n > 2
if tanks[n-1] == 0:
print (0)
return
for i in range( k ):
maxDiff = maxDiff + tanks[(n-1)-(i+1)]
print ( maxDiff )
return
numTest = int( input() )
while numTest:
numTest -= 1
codeforces1430ProblemB()
``` | instruction | 0 | 7,760 | 8 | 15,520 |
Yes | output | 1 | 7,760 | 8 | 15,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
T = int(input())
for testcase in range(T):
n,k = map(int,input().split())
a = list(map(int,input().split()))
a.sort()
for i in range(k):
if n-2-i == -1:
break
a[-1] += a[n-2-i]
a[n-2-i] = 0
print(max(a)-min(a))
``` | instruction | 0 | 7,761 | 8 | 15,522 |
Yes | output | 1 | 7,761 | 8 | 15,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
Submitted Solution:
```
for i in range(int(input())):
n,k=[int(num) for num in input().split()]
count=0
x=list(map(int,input().split()))
x.sort(reverse=True)
for i in range(k+1):
count=count+x[i]
print(count)
``` | instruction | 0 | 7,762 | 8 | 15,524 |
Yes | output | 1 | 7,762 | 8 | 15,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
Submitted Solution:
```
for _ in range(int(input())):
n,k = map(int,input().split())
capacities = [int(x) for x in input().split()]
capacities.sort()
if capacities[n - 1] == 0 or n == 1:
print(capacities[0])
else:
j = n - 2
for i in range(k):
if j >= 0:
capacities[n - 1] += capacities[j]
j -= 1
else:
break
print(capacities[n - 1] - capacities[0])
``` | instruction | 0 | 7,763 | 8 | 15,526 |
No | output | 1 | 7,763 | 8 | 15,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
Submitted Solution:
```
# your code goes here
from collections import *
t = int(input())
for i in range(t):
n, k = map(int, input().split())
L = list(map(int, input().split()))
L.sort()
print(sum(L[-1-k:-1]))
``` | instruction | 0 | 7,764 | 8 | 15,528 |
No | output | 1 | 7,764 | 8 | 15,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
Submitted Solution:
```
def calc(barrels):
i = barrels.index(max(barrels))
buf = barrels[i]
barrels[i] = '0'
j = barrels.index(max(barrels))
barrels[i] = buf
return i, j
n = int(input())
results = []
for i in range(0, n, 1):
nk = input().split()
barrels = input().split()
for k in range(0, int(nk[1]), 1):
res = calc(barrels)
barrels[res[0]] = int(barrels[res[0]]) + int(barrels[res[1]])
barrels[res[1]] = '0'
barrels[res[0]] = str(barrels[res[0]])
results.append(barrels[res[0]])
for element in results:
print(int(element))
``` | instruction | 0 | 7,765 | 8 | 15,530 |
No | output | 1 | 7,765 | 8 | 15,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n barrels lined up in a row, numbered from left to right from one. Initially, the i-th barrel contains a_i liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels x and y (the x-th barrel shouldn't be empty) and pour any possible amount of water from barrel x to barrel y (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Some examples:
* if you have four barrels, each containing 5 liters of water, and k = 1, you may pour 5 liters from the second barrel into the fourth, so the amounts of water in the barrels are [5, 0, 5, 10], and the difference between the maximum and the minimum is 10;
* if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still 0.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains two integers n and k (1 β€ k < n β€ 2 β
10^5) β the number of barrels and the number of pourings you can make.
The second line contains n integers a_1, a_2, ..., a_n (0 β€ a_i β€ 10^{9}), where a_i is the initial amount of water the i-th barrel has.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 β
10^5.
Output
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most k times.
Example
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
Submitted Solution:
```
t = int(input())
for _ in range(t):
k,n = tuple(map(int,input().split(' ')))
a = list(map(int,input().split(' ')))
for i in range(k):
a[1] += a[0]
a[0] = 0
a.sort()
print(a[len(a)-1]-a[0])
``` | instruction | 0 | 7,766 | 8 | 15,532 |
No | output | 1 | 7,766 | 8 | 15,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An infinitely long railway has a train consisting of n cars, numbered from 1 to n (the numbers of all the cars are distinct) and positioned in arbitrary order. David Blaine wants to sort the railway cars in the order of increasing numbers. In one move he can make one of the cars disappear from its place and teleport it either to the beginning of the train, or to the end of the train, at his desire. What is the minimum number of actions David Blaine needs to perform in order to sort the train?
Input
The first line of the input contains integer n (1 β€ n β€ 100 000) β the number of cars in the train.
The second line contains n integers pi (1 β€ pi β€ n, pi β pj if i β j) β the sequence of the numbers of the cars in the train.
Output
Print a single integer β the minimum number of actions needed to sort the railway cars.
Examples
Input
5
4 1 2 5 3
Output
2
Input
4
4 1 3 2
Output
2
Note
In the first sample you need first to teleport the 4-th car, and then the 5-th car to the end of the train. | instruction | 0 | 7,962 | 8 | 15,924 |
Tags: constructive algorithms, greedy
Correct Solution:
```
#!/usr/bin/env python3
# 606C_sort.py - Codeforces.com/problemset/problem/606/C by Sergey 2015
import unittest
import sys
###############################################################################
# Sort Class (Main Program)
###############################################################################
class Sort:
""" Sort representation """
def __init__(self, test_inputs=None):
""" Default constructor """
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
return next(it) if it else sys.stdin.readline().rstrip()
# Reading single elements
[self.n] = map(int, uinput().split())
# Reading a single line of multiple elements
self.nums = list(map(int, uinput().split()))
# Translate number to position
tr = [0] * self.n
for i in range(self.n):
tr[self.nums[i]-1] = i
# Longest increasing subarray
max_len = 1
for i in range(self.n):
if i > 0 and tr[i] > tr[i-1]:
cur_len += 1
max_len = max(max_len, cur_len)
else:
cur_len = 1
self.result = self.n - max_len
def calculate(self):
""" Main calcualtion function of the class """
return str(self.result)
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_single_test(self):
""" Sort class testing """
# Constructor test
test = "5\n4 1 2 5 3"
d = Sort(test)
self.assertEqual(d.n, 5)
self.assertEqual(d.nums, [4, 1, 2, 5, 3])
# Sample test
self.assertEqual(Sort(test).calculate(), "2")
# Sample test
test = "4\n4 1 3 2"
self.assertEqual(Sort(test).calculate(), "2")
# Sample test
test = "8\n6 2 1 8 5 7 3 4"
self.assertEqual(Sort(test).calculate(), "5")
# My tests
test = ""
# self.assertEqual(Sort(test).calculate(), "0")
# Time limit test
# self.time_limit_test(5000)
def time_limit_test(self, nmax):
""" Timelimit testing """
import random
import timeit
# Random inputs
test = str(nmax) + " " + str(nmax) + "\n"
numnums = [str(i) + " " + str(i+1) for i in range(nmax)]
test += "\n".join(numnums) + "\n"
nums = [random.randint(1, 10000) for i in range(nmax)]
test += " ".join(map(str, nums)) + "\n"
# Run the test
start = timeit.default_timer()
d = Sort(test)
calc = timeit.default_timer()
d.calculate()
stop = timeit.default_timer()
print("\nTimelimit Test: " +
"{0:.3f}s (init {1:.3f}s calc {2:.3f}s)".
format(stop-start, calc-start, stop-calc))
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
sys.stdout.write(Sort().calculate())
``` | output | 1 | 7,962 | 8 | 15,925 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are the top spy of AtCoder Kingdom. To prevent the stolen secret from being handed to AlDebaran Kingdom, you have sneaked into the party where the transaction happens.
There are N attendees in the party, and they are given attendee numbers from 1 through N. The height of Attendee i is A_i.
According to an examination beforehand, you know that a pair of attendees satisfying the condition below will make the transaction.
* The absolute difference of their attendee numbers is equal to the sum of their heights.
There are \frac{N(N-1)}{2} ways to choose two from the N attendees and make a pair. Among them, how many satisfy the condition above?
P.S.: We cannot let you know the secret.
Constraints
* All values in input are integers.
* 2 \leq N \leq 2 \times 10^5
* 1 \leq A_i \leq 10^9\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 \dots A_N
Output
Print the number of pairs satisfying the condition.
Examples
Input
6
2 3 3 1 3 1
Output
3
Input
6
5 2 4 2 8 8
Output
0
Input
32
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5
Output
22
Submitted Solution:
```
num = int(input())
height_list = [0] # 1ε§γΎγγ§εγγγγγ
height_list.extend(list(map(int, input().split())))
counta = 0
for i in range(1, num+1):
for j in range(i, num+1):
if abs(i-j) == height_list[i] + height_list[j]:
counta += 1
print(counta)
``` | instruction | 0 | 8,198 | 8 | 16,396 |
No | output | 1 | 8,198 | 8 | 16,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again. | instruction | 0 | 8,530 | 8 | 17,060 |
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
import sys
import collections as cc
import bisect as bi
I=lambda:list(map(int,input().split()))
cur=1
def f(x):
return x*(x+1)//2
ar=[f((2**i)-1) for i in range(1,32)]
for i in range(1,len(ar)):
ar[i]+=ar[i-1]
for tc in range(int(input())):
n,=I()
temp=bi.bisect_right(ar,n)
print(temp)
``` | output | 1 | 8,530 | 8 | 17,061 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again. | instruction | 0 | 8,531 | 8 | 17,062 |
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
def sum_ab(a,b):
return int((a + b) / 2 * (b - a + 1))
for _ in range(int((input()))):
schody = int(input())
koszt = 1
cos = 1
pot = 2
x = 0
while schody >= koszt:
schody -= koszt
cos += pot
pot *= 2
koszt = sum_ab(1, cos)
x += 1
print(x)
``` | output | 1 | 8,531 | 8 | 17,063 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again. | instruction | 0 | 8,532 | 8 | 17,064 |
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
n = 10
ans = [1]
for i in range(1, 32):
ans.append(ans[-1]*2+(2**i)**2)
for i in range(1, len(ans)):
ans[i] = ans[i]+ans[i-1]
t = int(input())
for _ in range(t):
num = int(input())
j = 0
while num >= ans[j]:
j = j + 1
print(j)
# print(ans[-1])
# for _ in range(t):
``` | output | 1 | 8,532 | 8 | 17,065 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again. | instruction | 0 | 8,533 | 8 | 17,066 |
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
x=int(input())
i,curr=0,0
while True:
curr = 2*curr+1
val = (curr*(curr+1))//2
if x>=val:
x-=val
i+=1
else:break
print(i)
``` | output | 1 | 8,533 | 8 | 17,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again. | instruction | 0 | 8,534 | 8 | 17,068 |
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
ar=[]
r=1
while(1):
x=(r*(r+1))//2
if(x>1e20):
break
ar.append(x)
r=r*2+1
t=int(input())
for a0 in range(t):
x=int(input())
r=0;
while(x>0):
r+=1
x-=ar[r]
print(r)
``` | output | 1 | 8,534 | 8 | 17,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again. | instruction | 0 | 8,535 | 8 | 17,070 |
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
def main():
n = int(input())
m = 0
k = 2
while n >= 0:
n -= (k - 1) * ((k - 1)// 2 + 1)
k *= 2
if n < 0:
break
m += 1
print(m)
if __name__ == '__main__':
t = int(input())
for i in range(t):
main()
``` | output | 1 | 8,535 | 8 | 17,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again. | instruction | 0 | 8,536 | 8 | 17,072 |
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
import math
t = int(input())
while (t):
x = int(input())
cnt = 0
i = 1
while (True):
if (i*(i+1)//2 <= x):
cnt += 1
else:
break
x -= i*(i+1)//2
i = i*2 + 1
print(cnt)
t -= 1
``` | output | 1 | 8,536 | 8 | 17,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again. | instruction | 0 | 8,537 | 8 | 17,074 |
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c for j in range(b)] for i in range(a)]
def list3d(a, b, c, d): return [[[d for k in range(c)] for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e for l in range(d)] for k in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10**19
MOD = 10**9 + 7
EPS = 10**-10
def rec(sm, cnt, h, x):
if sm > x:
return 0
return rec(sm+cnt*2+(h+1)**2, cnt*2+(h+1)**2, h*2+1, x) + 1
for _ in range(INT()):
x = INT()
ans = rec(1, 1, 1, x)
print(ans)
``` | output | 1 | 8,537 | 8 | 17,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
n = int(input())
p = 1
s = 0
ans = 0
while True:
p *= 2
s += (p - 1) * p // 2
if s > n:
break
ans += 1
print(ans)
main()
``` | instruction | 0 | 8,538 | 8 | 17,076 |
Yes | output | 1 | 8,538 | 8 | 17,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
for i in range(int(input())):
x=int(input())
count=0
stair=1
cell=1
while((x-cell)>=0):
x=x-cell
count+=1
stair=(stair*2)+1
cell=(stair*(stair+1))//2
print(count)
``` | instruction | 0 | 8,539 | 8 | 17,078 |
Yes | output | 1 | 8,539 | 8 | 17,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
import math
t = int(input())
for _ in range(t):
n = int(input())
a = 1
tot = 1
cost = 1
ans = 1
while n>=tot:
a <<= 1
cost = 2*cost + a*a
tot += cost
ans += 1
print(ans-1)
``` | instruction | 0 | 8,540 | 8 | 17,080 |
Yes | output | 1 | 8,540 | 8 | 17,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
q = int(input())
for _ in range(q):
n = int(input())
wyn = 0
pot = 1
total = 1
while total <= n:
wyn += 1
pot += 1
total += (2**pot-1)*(2**pot)//2
print(wyn)
``` | instruction | 0 | 8,541 | 8 | 17,082 |
Yes | output | 1 | 8,541 | 8 | 17,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
from math import *
a = int(input())
for x in range(a):
q = int(input())
if q == 1:
print(1)
else:
print(floor(log(q,4)))
``` | instruction | 0 | 8,542 | 8 | 17,084 |
No | output | 1 | 8,542 | 8 | 17,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
arr=[]
for i in range(1,32):
arr.append(2**i-1)
temp=[]
for i in arr:
temp.append(i*(i+1)/2)
res=[temp[0]]
for i in range(1,len(temp)):
res.append(res[-1]+temp[i])
for _ in range(int(input())):
n=int(input())
if n<=6:
print(1)
elif 7<=n<=34:
print(2)
elif 35<=n<=154:
print(3)
else:
for i in range(len(res)):
if n<res[i]:
print(i)
break
``` | instruction | 0 | 8,543 | 8 | 17,086 |
No | output | 1 | 8,543 | 8 | 17,087 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.