message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1].
Submitted Solution:
```
import io,os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n = int(input())
a = [int(x) - 1 for x in input().split()]
l = [-1 for _ in range(n + 1)]
r = [-1 for _ in range(n + 1)]
freq = [0 for _ in range(n + 1)]
dp = [0 for _ in range(n + 1)]
for i in range(n):
if(l[a[i]] == -1): l[a[i]] = i
r[a[i]] = i
for i in range(n - 1, -1, -1):
dp[i] = dp[i + 1]
freq[a[i]] += 1
if (i == l[a[i]]): dp[i] = max(dp[i], dp[r[a[i]] + 1] + freq[a[i]])
else : dp[i] = max(dp[i], freq[a[i]])
print(n - dp[0])
``` | instruction | 0 | 103,118 | 8 | 206,236 |
Yes | output | 1 | 103,118 | 8 | 206,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1].
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import Counter
n = int(input())
A = list(map(int, input().split()))
cnt = Counter(A)
left, right = {}, {}
for i, a in enumerate(A):
left.setdefault(a, i)
right[a] = i
dp = [0] * (n + 1)
cnt2 = Counter()
for i in range(n - 1, -1, -1):
dp[i] = dp[i + 1]
cnt2[A[i]] += 1
if i == left[A[i]]:
dp[i] = max(dp[i], dp[right[A[i]] + 1] + cnt[A[i]])
else:
dp[i] = max(dp[i], cnt2[A[i]])
print(n - dp[0])
``` | instruction | 0 | 103,119 | 8 | 206,238 |
Yes | output | 1 | 103,119 | 8 | 206,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1].
Submitted Solution:
```
def show_key(book):
c_book = book
keys = []
s = 0
for key, i in enumerate(c_book):
if i!=0:
keys.append([])
for key1, i1 in enumerate(c_book):
if i == i1:
keys[s].append(key1)
c_book[key1] = 0
s += 1
return keys
def search_key(keys, key):
for i in keys:
for key1, j in enumerate(i):
if j==key:
return keys.index(i), key1
def if_sorted(mas):
mas.sort()
for key, i in enumerate(mas):
if key!=len(mas)-1:
if abs(i - mas[key+1])!=1:
return False
return True
def move_books(keys, key):
for i in keys:
for key1, j in enumerate(i):
if j==key:
i[key1] = len(book)-1
if j>key and j!=key:
i[key1]-=1
return keys
def find_min(keys):
min = 9999
for i in keys:
if len(i)>1 and not if_sorted(i) and len(i) < min:
min = len(i)
out = i
if min == 9999:
return False
else:
return out
def if_keys_sorted(keys):
for i in keys:
if not if_sorted(i):
return False
return True
_ = input()
book = input().split()
book = [int(n) for n in book]
keys = show_key(book)
s = 0
for i in keys:
if i == find_min(keys) and not if_sorted(i) and not if_keys_sorted(keys):
for j in i:
if not if_sorted(i) and not if_keys_sorted(keys):
move_books(keys, j)
s+=1
else:
break
print(s)
``` | instruction | 0 | 103,120 | 8 | 206,240 |
No | output | 1 | 103,120 | 8 | 206,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1].
Submitted Solution:
```
from collections import Counter
n = int(input())
books = input().split()
act = 0
co = dict(Counter(books))
#print(co)
onln = 1
for i in range(0, len(books) - 1):
if books[i] == books[-1]:
if books[i] != books[i + 1]:
act += onln
#print(books)
books = books[: i - onln + 1] + books[i + 1: ] + books[i - onln + 1: i + 1]
#print(books)
onln = 1
else:
onln += 1
#print(act)
onln = 1
for i in range(0, len(books) - 1):
if books[i] != books[i + 1]:
#print(books[i], '-', i)
if onln != co[books[i]]:
act += onln
onln = 1
else:
onln += 1
print(act)
``` | instruction | 0 | 103,121 | 8 | 206,242 |
No | output | 1 | 103,121 | 8 | 206,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1].
Submitted Solution:
```
def bookself(a):
count=0
for i in range(len(a)-1,-1,-1):
key=i
for j in range(i-1,-1,-1):
if(a[key]>a[j]):
key=j
if key!=i:
count+=1
a[i],a[key]=a[key],a[i]
return count
n=input()
s=list(map(int,input().split(" ")))
print(bookself(s))
``` | instruction | 0 | 103,122 | 8 | 206,244 |
No | output | 1 | 103,122 | 8 | 206,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day you wanted to read something, so you went to your bookshelf to grab some book. But when you saw how messy the bookshelf was you decided to clean it up first.
<image>
There are n books standing in a row on the shelf, the i-th book has color a_i.
You'd like to rearrange the books to make the shelf look beautiful. The shelf is considered beautiful if all books of the same color are next to each other.
In one operation you can take one book from any position on the shelf and move it to the right end of the shelf.
What is the minimum number of operations you need to make the shelf beautiful?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of books.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the book colors.
Output
Output the minimum number of operations to make the shelf beautiful.
Examples
Input
5
1 2 2 1 3
Output
2
Input
5
1 2 2 1 1
Output
1
Note
In the first example, we have the bookshelf [1, 2, 2, 1, 3] and can, for example:
1. take a book on position 4 and move to the right end: we'll get [1, 2, 2, 3, 1];
2. take a book on position 1 and move to the right end: we'll get [2, 2, 3, 1, 1].
In the second example, we can move the first book to the end of the bookshelf and get [2,2,1,1,1].
Submitted Solution:
```
from collections import defaultdict, deque, Counter
# t = int(input())
t = 1
while t:
t-=1
n= int(input())
arr = [int(x) for x in input().split()]
for i in range(n-1,0,-1):
if arr[i] == arr[i-1]:
arr.pop()
else:
arr.pop()
break
c = Counter(arr)
maxx = -1
for key in c:
if c[key] > maxx:
remove = key
maxx = c[key]
res = len(arr) - maxx
print(res)
``` | instruction | 0 | 103,123 | 8 | 206,246 |
No | output | 1 | 103,123 | 8 | 206,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's been long after the events of the previous problems, and Karen has now moved on from student life and is looking to relocate to a new neighborhood.
<image>
The neighborhood consists of n houses in a straight line, labelled 1 to n from left to right, all an equal distance apart.
Everyone in this neighborhood loves peace and quiet. Because of this, whenever a new person moves into the neighborhood, he or she always chooses the house whose minimum distance to any occupied house is maximized. If there are multiple houses with the maximum possible minimum distance, he or she chooses the leftmost one.
Note that the first person to arrive always moves into house 1.
Karen is the k-th person to enter this neighborhood. If everyone, including herself, follows this rule, which house will she move into?
Input
The first and only line of input contains two integers, n and k (1 ≤ k ≤ n ≤ 1018), describing the number of houses in the neighborhood, and that Karen was the k-th person to move in, respectively.
Output
Output a single integer on a line by itself, the label of the house Karen will move into.
Examples
Input
6 4
Output
2
Input
39 3
Output
20
Note
In the first test case, there are 6 houses in the neighborhood, and Karen is the fourth person to move in:
1. The first person moves into house 1.
2. The second person moves into house 6.
3. The third person moves into house 3.
4. The fourth person moves into house 2.
In the second test case, there are 39 houses in the neighborhood, and Karen is the third person to move in:
1. The first person moves into house 1.
2. The second person moves into house 39.
3. The third person moves into house 20.
Submitted Solution:
```
dict1 = {}
a = input("enter two integers separated by space:")
c, d = a.split()
n = int(c)
k = int(d)
if k == 1:
print(1)
for i in range(1, n+1):
dict1[i] = ""
dict1[1] = "occupied"
for i in range(2, k+1):
ans1 = []
minDistDict = {}
occupiedList = []
for j in range(1, n+1):
if dict1[j] == "occupied":
occupiedList.append(j)
availableList = [x for x in range(1, n+1) if x not in occupiedList]
for l in availableList:
distList = []
for o in occupiedList:
distList.append(abs(l-o))
minDistDict[l] = min(distList)
for key in minDistDict:
if minDistDict[key] == max(list(minDistDict.values())):
ans1.append(key)
dict1[min(ans1)] = "occupied"
if i == k:
print(min(ans1))
``` | instruction | 0 | 103,379 | 8 | 206,758 |
No | output | 1 | 103,379 | 8 | 206,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's been long after the events of the previous problems, and Karen has now moved on from student life and is looking to relocate to a new neighborhood.
<image>
The neighborhood consists of n houses in a straight line, labelled 1 to n from left to right, all an equal distance apart.
Everyone in this neighborhood loves peace and quiet. Because of this, whenever a new person moves into the neighborhood, he or she always chooses the house whose minimum distance to any occupied house is maximized. If there are multiple houses with the maximum possible minimum distance, he or she chooses the leftmost one.
Note that the first person to arrive always moves into house 1.
Karen is the k-th person to enter this neighborhood. If everyone, including herself, follows this rule, which house will she move into?
Input
The first and only line of input contains two integers, n and k (1 ≤ k ≤ n ≤ 1018), describing the number of houses in the neighborhood, and that Karen was the k-th person to move in, respectively.
Output
Output a single integer on a line by itself, the label of the house Karen will move into.
Examples
Input
6 4
Output
2
Input
39 3
Output
20
Note
In the first test case, there are 6 houses in the neighborhood, and Karen is the fourth person to move in:
1. The first person moves into house 1.
2. The second person moves into house 6.
3. The third person moves into house 3.
4. The fourth person moves into house 2.
In the second test case, there are 39 houses in the neighborhood, and Karen is the third person to move in:
1. The first person moves into house 1.
2. The second person moves into house 39.
3. The third person moves into house 20.
Submitted Solution:
```
from sys import stdin
from collections import deque
def main():
t = list(stdin.readline())
b = False
n = ""
k = ""
for i in t:
if i == " " :
b = True
if b == False :
n += i
else :
k += i
n = int(n)
k = int(k)
x = deque()
answer = 0
m = int(n / 2)
x.append(n)
x.append(0)
if k > 2:
k = k - 2
while k > 0 :
temp = x.pop()
if temp == n :
m = int(m/2)
x.appendleft(n)
continue
x.appendleft(temp)
x.appendleft(temp + m)
answer = temp + m
k = k - 1
print(answer + 1)
elif k == 1:
print(1)
else:
print(n)
main()
``` | instruction | 0 | 103,380 | 8 | 206,760 |
No | output | 1 | 103,380 | 8 | 206,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's been long after the events of the previous problems, and Karen has now moved on from student life and is looking to relocate to a new neighborhood.
<image>
The neighborhood consists of n houses in a straight line, labelled 1 to n from left to right, all an equal distance apart.
Everyone in this neighborhood loves peace and quiet. Because of this, whenever a new person moves into the neighborhood, he or she always chooses the house whose minimum distance to any occupied house is maximized. If there are multiple houses with the maximum possible minimum distance, he or she chooses the leftmost one.
Note that the first person to arrive always moves into house 1.
Karen is the k-th person to enter this neighborhood. If everyone, including herself, follows this rule, which house will she move into?
Input
The first and only line of input contains two integers, n and k (1 ≤ k ≤ n ≤ 1018), describing the number of houses in the neighborhood, and that Karen was the k-th person to move in, respectively.
Output
Output a single integer on a line by itself, the label of the house Karen will move into.
Examples
Input
6 4
Output
2
Input
39 3
Output
20
Note
In the first test case, there are 6 houses in the neighborhood, and Karen is the fourth person to move in:
1. The first person moves into house 1.
2. The second person moves into house 6.
3. The third person moves into house 3.
4. The fourth person moves into house 2.
In the second test case, there are 39 houses in the neighborhood, and Karen is the third person to move in:
1. The first person moves into house 1.
2. The second person moves into house 39.
3. The third person moves into house 20.
Submitted Solution:
```
from sys import stdin
from collections import deque
def main():
t = list(stdin.readline())
b = False
n = ""
k = ""
for i in t:
if i == " " :
b = True
if b == False :
n += i
else :
k += i
n = int(n)
k = int(k)
x = deque()
answer = 0
m = int(n / 2)
x.append(n)
x.append(0)
k = k - 2
while k > 0 :
temp = x.pop()
if temp == n :
m = int(m/2)
x.appendleft(n)
continue
x.appendleft(temp)
x.appendleft(temp + m)
answer = temp + m
k = k - 1
if answer == 0:
print(n)
else:
print(answer + 1)
main()
``` | instruction | 0 | 103,381 | 8 | 206,762 |
No | output | 1 | 103,381 | 8 | 206,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's been long after the events of the previous problems, and Karen has now moved on from student life and is looking to relocate to a new neighborhood.
<image>
The neighborhood consists of n houses in a straight line, labelled 1 to n from left to right, all an equal distance apart.
Everyone in this neighborhood loves peace and quiet. Because of this, whenever a new person moves into the neighborhood, he or she always chooses the house whose minimum distance to any occupied house is maximized. If there are multiple houses with the maximum possible minimum distance, he or she chooses the leftmost one.
Note that the first person to arrive always moves into house 1.
Karen is the k-th person to enter this neighborhood. If everyone, including herself, follows this rule, which house will she move into?
Input
The first and only line of input contains two integers, n and k (1 ≤ k ≤ n ≤ 1018), describing the number of houses in the neighborhood, and that Karen was the k-th person to move in, respectively.
Output
Output a single integer on a line by itself, the label of the house Karen will move into.
Examples
Input
6 4
Output
2
Input
39 3
Output
20
Note
In the first test case, there are 6 houses in the neighborhood, and Karen is the fourth person to move in:
1. The first person moves into house 1.
2. The second person moves into house 6.
3. The third person moves into house 3.
4. The fourth person moves into house 2.
In the second test case, there are 39 houses in the neighborhood, and Karen is the third person to move in:
1. The first person moves into house 1.
2. The second person moves into house 39.
3. The third person moves into house 20.
Submitted Solution:
```
from sys import stdin
from collections import deque
def main():
t = list(stdin.readline())
b = False
n = ""
k = ""
for i in t:
if i == " " :
b = True
if b == False :
n += i
else :
k += i
n = int(n)
k = int(k)
x = deque()
answer = 0
m = int(n / 2)
x.append(n)
x.append(0)
k = k - 2
while k > 0 :
temp = x.pop()
if temp == n :
m = int(m/2)
x.appendleft(n)
continue
x.appendleft(temp)
x.appendleft(temp + m)
answer = temp + m
k = k - 1
print(answer + 1)
main()
``` | instruction | 0 | 103,382 | 8 | 206,764 |
No | output | 1 | 103,382 | 8 | 206,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21. | instruction | 0 | 103,665 | 8 | 207,330 |
Tags: implementation
Correct Solution:
```
# import sys
# sys.stdin = open("test.in","r")
# sys.stdout = open("test.out","w")
n,d=map(int,input().split())
a=list(map(int,input().split()))
c=0
for i in range(n-1):
if a[i+1]-a[i]>2*d:
c+=2
elif a[i+1]-a[i]==2*d:
c+=1
print(c+2)
``` | output | 1 | 103,665 | 8 | 207,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21. | instruction | 0 | 103,668 | 8 | 207,336 |
Tags: implementation
Correct Solution:
```
n,d=map(int,input().split())
a=list(map(int,input().split()))
ans=0
for i in range(n-1):
if a[i]+d<=a[i+1]-d:
ans+=1
if a[i+1]-d>a[i]+d:
ans+=1
ans+=2
print(ans)
``` | output | 1 | 103,668 | 8 | 207,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21. | instruction | 0 | 103,669 | 8 | 207,338 |
Tags: implementation
Correct Solution:
```
s=input().split()
n=int(s[0])
d=int(s[1])
t=input().split()
ans=0
ans=ans+2
l=len(t)
for i in range(l):
t[i]=int(t[i])
q=0
d2=2*d
for i in range(1,l):
q=t[i]-t[i-1]
if q > d2:
ans+=2
elif q == d2:
ans+=1
print(ans)
``` | output | 1 | 103,669 | 8 | 207,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21. | instruction | 0 | 103,671 | 8 | 207,342 |
Tags: implementation
Correct Solution:
```
n,k=map(int,input().split())
l=[int(i) for i in input().split()]
c=2
for i in range(1,n):
if l[i]-l[i-1]>2*k:
c+=2
elif l[i]-l[i-1]==2*k:
c+=1
print(c)
``` | output | 1 | 103,671 | 8 | 207,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21. | instruction | 0 | 103,672 | 8 | 207,344 |
Tags: implementation
Correct Solution:
```
input1 = input()
n = int(input1.split()[0])
d = int(input1.split()[1])
coor = [int(num) for num in input().split()]
sum = 0
for i in range(n-1):
if(coor[i+1]-coor[i] == 2*d):
sum = sum+1
if(coor[i+1]-coor[i] > 2*d):
sum = sum+2
print(sum+2)
``` | output | 1 | 103,672 | 8 | 207,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21.
Submitted Solution:
```
n,q=map(int,input().split())
arr=list(map(int,input().split()))
ct=0
arr.sort()
for i in range(1,n):
dif=arr[i]-arr[i-1]
if (dif)/q==2:
ct+=1
elif dif/q>2:
ct+=2
#print(dif,ct)
print(ct+2)
``` | instruction | 0 | 103,673 | 8 | 207,346 |
Yes | output | 1 | 103,673 | 8 | 207,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21.
Submitted Solution:
```
n, d = map(int, input().split())
x = list(map(int, input().split()))
l, r = [x[i] - d for i in range(n)], [x[i] + d for i in range(n)]
lr = set(l) | set(r)
ans = 0
for lri in lr:
if min(list(map(lambda y: abs(lri - y), x))) == d:
ans += 1
print(ans)
``` | instruction | 0 | 103,674 | 8 | 207,348 |
Yes | output | 1 | 103,674 | 8 | 207,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21.
Submitted Solution:
```
m,n = map(int,input().split())
a = list(map(int,input().split()))
a.sort()
count = 2
for i in range(m-1):
if a[i+1]-a[i] > 2*n:
count+=2
elif a[i+1]-a[i] == 2*n:
count+=1
print(count)
``` | instruction | 0 | 103,675 | 8 | 207,350 |
Yes | output | 1 | 103,675 | 8 | 207,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21.
Submitted Solution:
```
def input_int():
return list(map(int, input().split()))
n, k = input_int()
arr = input_int()
from collections import defaultdict
vis = defaultdict(int)
ans = 2
for i in range(1, n):
for d in [-1, 1]:
if d == -1:
cur = arr[i-1] + k
else:
cur = arr[i] - k
if abs(cur - arr[i-1]) >= k and abs(cur - arr[i]) >= k:
if vis[cur] == 0:
vis[cur] = 1
ans += 1
print(ans)
``` | instruction | 0 | 103,676 | 8 | 207,352 |
Yes | output | 1 | 103,676 | 8 | 207,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21.
Submitted Solution:
```
# cook your dish here
n,d=input().split()
n=int(n)
d=int(d)
a=[int(n) for n in input().split()]
o=[]
d=0
for i in range(len(a)-1):
o.append(a[i+1]-a[i])
for j in range(len(o)):
if o[j]>=2*d:
d+=o[j]-(2*d)+1
print(d)
``` | instruction | 0 | 103,677 | 8 | 207,354 |
No | output | 1 | 103,677 | 8 | 207,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21.
Submitted Solution:
```
q, w = map(int, input().split())
e = list(map(int, input().split()))
d = 0
for i in range(1, q - 1):
if e[i + 1] - (e[i] + w) > w:
d += 1
if (e[i] + w) - e[i - 1] > w:
d += 1
#if e[1] - e[0] + w > w:
# d += 1
print(d + 2)
``` | instruction | 0 | 103,678 | 8 | 207,356 |
No | output | 1 | 103,678 | 8 | 207,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21.
Submitted Solution:
```
n, d = map(int, input().split())
a = list(map(int, input().split()))
hotels = 2
if n == 1:
print(hotels)
else:
for i in range(n - 1):
dis = (a[i + 1] - a[i]) // 2
if dis >= d:
l = a[i] + dis
r = a[i + 1] - dis
hotels += len(list(range(l, r + 1)))
print(hotels)
``` | instruction | 0 | 103,679 | 8 | 207,358 |
No | output | 1 | 103,679 | 8 | 207,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has n hotels, where the i-th hotel is located in the city with coordinate x_i. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to d. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original n hotels to the new one is equal to d.
Input
The first line contains two integers n and d (1≤ n≤ 100, 1≤ d≤ 10^9) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains n different integers in strictly increasing order x_1, x_2, …, x_n (-10^9≤ x_i≤ 10^9) — coordinates of Sonya's hotels.
Output
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to d.
Examples
Input
4 3
-3 2 9 16
Output
6
Input
5 2
4 8 11 18 19
Output
5
Note
In the first example, there are 6 possible cities where Sonya can build a hotel. These cities have coordinates -6, 5, 6, 12, 13, and 19.
In the second example, there are 5 possible cities where Sonya can build a hotel. These cities have coordinates 2, 6, 13, 16, and 21.
Submitted Solution:
```
q, w = map(int, input().split())
e = list(map(int, input().split()))
d = 0
v = 0
for i in range(1, q - 1):
if e[i + 1] - (e[i] + w) >= w and e[i] + w <= e[i + 1]:
d += 1
if (e[i] - w) - e[i - 1] >= w:
d += 1
if e[len(e) - 1] - w - e[len(e) - 2] >= w:
d += 1
if e[1] - (e[0] + w) >= w:
d += 1
for i in range(1, q - 1):
if e[i] - w - e[i - 1] == w:
d -= 1
if e[i + 1] - (e[i] + w) == w:
d -= 1
print(d + 2)
``` | instruction | 0 | 103,680 | 8 | 207,360 |
No | output | 1 | 103,680 | 8 | 207,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15. | instruction | 0 | 103,789 | 8 | 207,578 |
Tags: data structures, dp, greedy
Correct Solution:
```
def main():
while True:
try:
n, m = map(int, input().strip().split())
s = str(input())
print(getAns(n, m, s))
except EOFError:
break
def getAns(n, k, s):
ans = [0] * (n + 10)
s = '0' + s
ans[0] = 0
lrt = 0
for i in range(1, n+1, 1):
while (lrt < i and (lrt < i - k or s[lrt]=='0')):
lrt += 1
# print('nb', s[lrt], i, lrt)
ans[i] = ans[max(0, lrt-k-1)]+lrt if s[lrt]=='1' else ans[i-1]+i
if s[i] == '1':
for j in range(i-1, -1, -1):
if s[j] == '1':
break
ans[j] = min(ans[j], ans[i])
# print(ans)
return ans[n]
if __name__ == '__main__':
main()
``` | output | 1 | 103,789 | 8 | 207,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15. | instruction | 0 | 103,790 | 8 | 207,580 |
Tags: data structures, dp, greedy
Correct Solution:
```
def find(roomcount,radius,string):
zero_roomcount=[0]*(roomcount+1)
binary_move=1<<1000#сдвиг двоичной единицы на 1000 влево или любое число
for i in range(roomcount,0,-1):
if string[i-1]=='1':
binary_move=i
zero_roomcount[i]=binary_move
dp=[0]
for i in range(1,roomcount+1):
dp.append(dp[-1]+i)
c=zero_roomcount[max(i-radius,1)]
if c<=i+radius:
dp[i]=min(dp[i],dp[max(1,c-radius)-1]+c)
return dp[roomcount]
roomcount,radius=map(int,input().split())
string=input()
print(find(roomcount,radius,string))
``` | output | 1 | 103,790 | 8 | 207,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15. | instruction | 0 | 103,791 | 8 | 207,582 |
Tags: data structures, dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
n,k=map(int,input().split())
s=input().strip()
seg_el=1<<((n+k+1).bit_length())# Segment treeの台の要素数
SEG=[1<<40]*(2*seg_el)# 1-indexedなので、要素数2*seg_el.Segment treeの初期値で初期化
def getvalue(n,seg_el):
i=n+seg_el
ANS=1<<40
ANS=min(SEG[i],ANS)
i>>=1# 子ノードへ
while i!=0:
ANS=min(SEG[i],ANS)
i>>=1
return ANS
def updates(l,r,x):# 区間[l,r)に関するminを調べる
L=l+seg_el
R=r+seg_el
while L<R:
if L & 1:
SEG[L]=min(x,SEG[L])
L+=1
if R & 1:
R-=1
SEG[R]=min(x,SEG[R])
L>>=1
R>>=1
updates(n,n+k+1,0)
for i in range(n-1,-1,-1):
if s[i]=="0":
x=getvalue(i+1,seg_el)
updates(i,i+1,x+i+1)
else:
x=getvalue(i+k+1,seg_el)
updates(max(0,i-k),i+k+1,x+i+1)
print(getvalue(0,seg_el))
``` | output | 1 | 103,791 | 8 | 207,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15. | instruction | 0 | 103,792 | 8 | 207,584 |
Tags: data structures, dp, greedy
Correct Solution:
```
import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
ii=lambda:int(input())
kk=lambda:map(int,input().split())
ll=lambda:list(kk())
n,k=kk()
s = input()
curr = n+k+1
lmr = [0]*n
for i in range(n-1,-k-1,-1):
if i >-1 and s[i]==49: curr=i
if (i+k<n): lmr[i+k] = curr
dp = [n*n]*(n+1)
dp[n]=0
for i in range(n, 0,-1):
dp[i-1]=min(dp[i-1], dp[i]+i)
if lmr[i-1]-(i-1)<=k:
lm = max(0, lmr[i-1]-k)
dp[lm]=min(dp[lm], dp[i]+lmr[i-1]+1)
print(dp[0])
``` | output | 1 | 103,792 | 8 | 207,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15. | instruction | 0 | 103,793 | 8 | 207,586 |
Tags: data structures, dp, greedy
Correct Solution:
```
if __name__ == '__main__':
n, dist = map(int, input().split())
light = input().strip()
right_cover = 0
left_most = [-1]*(n)
for idx, ele in enumerate(light):
if ele == '1':
for i in range(max(right_cover, idx-dist), min(idx + dist + 1, n)):
left_most[i] = idx
right_cover = idx + dist + 1
dp = [0]*(2*n+1)
for i in range(n):
dp[i] = 1000000000000000000
for idx in range(n-1, -1, -1):
dp[idx] = min(dp[idx+1] + idx + 1, dp[idx])
if left_most[idx] != -1:
left = left_most[idx]
dp[max(left-dist, 0)] = min(dp[idx+1] + left + 1, dp[max(left-dist, 0)])
dp[idx] = min(dp[max(left-dist, 0)], dp[idx])
print(dp[0])
``` | output | 1 | 103,793 | 8 | 207,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15. | instruction | 0 | 103,794 | 8 | 207,588 |
Tags: data structures, dp, greedy
Correct Solution:
```
n, k = tuple(map(int, input().rstrip().split()))
s = input().rstrip()
cur = int((2*pow(10,5)) * (2*pow(10,5) + 1) / 2) + 1
cost = [0] * (n+1)
dp = [0]
for i in range(n, 0, -1):
if s[i-1] == '1':
cur=i
cost[i]=cur
for i in range(1, n+1):
dp.append(dp[-1] + i)
cs = cost[max(i-k, 1)]
if cs <= i+k:
dp[i] = min(dp[i], dp[max(1, cs-k)-1]+cs)
print(dp[n])
``` | output | 1 | 103,794 | 8 | 207,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15. | instruction | 0 | 103,795 | 8 | 207,590 |
Tags: data structures, dp, greedy
Correct Solution:
```
n, k = map(int, input().split())
mask = list(map(int, input()))
dp = [0] * (n + 2)
nxt = [1 << 31] * (n + 2)
for i in range(n, 0, -1):
nxt[i] = i if mask[i - 1] is 1 else nxt[i + 1]
for i in range(1, n + 1):
dp[i] = dp[i - 1] + i
idx = nxt[max(1, i - k)]
if idx <= i + k:
dp[i] = min(dp[i], dp[max(0, idx - k - 1)] + idx)
print(dp[n])
``` | output | 1 | 103,795 | 8 | 207,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15. | instruction | 0 | 103,796 | 8 | 207,592 |
Tags: data structures, dp, greedy
Correct Solution:
```
import heapq
n, k = list(map(int, input().split()))
s = input()
mostRecent = n
best = []
for room in range(n-1, -1, -1):
if s[room] == '1':
mostRecent = room
best.append(mostRecent)
best = best[::-1]
dp = [0]
vals = [(0,0)]
for room in range(1, n + 1):
new = dp[-1] + room
if room - k - 1 >= 0:
bestRout = best[room - k - 1]
if bestRout <= (room - 1):
covered = bestRout - k
if covered >= 0:
try:
while len(vals) > 0 and vals[0][1] < covered:
heapq.heappop(vals)
if len(vals) > 0:
add = vals[0][0]
new2 = (bestRout + 1) + add
new = min(new2, new)
except Exception:
pass
else:
new2 = (bestRout + 1)
new = min(new2, new)
dp.append(new)
heapq.heappush(vals, (new, room))
print(new)
``` | output | 1 | 103,796 | 8 | 207,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15.
Submitted Solution:
```
from heapq import heappush, heappop
from collections import deque
N, K = map(int, input().split())
*B, = map(int, input())
INF = 10**18
que = deque()
hq = []
dp = [INF]*(N+1)
dp[0] = 0
for i in range(N):
a = dp[i]
while hq and hq[0][1] < i:
heappop(hq)
if hq:
a = min(hq[0][0], a)
#print(i, a)
while que and a <= que[-1][1]:
que.pop()
que.append((i, a))
if B[i]:
heappush(hq, (que[0][1] + (i+1), i+K+1))
if que and que[0][0] <= i-K:
que.popleft()
dp[i+1] = a + (i+1)
while hq and hq[0][1] < N:
heappop(hq)
a = dp[N]
if hq:
a = min(hq[0][0], a)
print(a)
``` | instruction | 0 | 103,797 | 8 | 207,594 |
Yes | output | 1 | 103,797 | 8 | 207,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15.
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(str, input().strip()))
ans = 0
import math
from collections import deque
dp = [math.inf] * (n + 10)
dp[n + 1] = 0
next = [n + 1] * (n + 10)
now = deque()
for i in range(n):
if a[i] == '1':
now.append(i + 1)
if len(now) == 0:
continue
if now[0] + k < i + 1:
now.popleft()
if len(now):
next[i + 1] = now[0]
cur = n + 1
for i in range(n - 1, -1, -1):
if a[i] == '1':
cur = i + 1
if cur - i - 1 <= k:
next[i + 1] = min(next[i + 1], cur)
for i in range(n, 0, -1):
dp[i] = min(dp[i], dp[i + 1] + i)
if next[i] != n + 1:
dp[max(next[i] - k, 1)] = min(dp[max(next[i] - k, 1)], dp[i + 1] + next[i])
print(dp[1])
``` | instruction | 0 | 103,798 | 8 | 207,596 |
Yes | output | 1 | 103,798 | 8 | 207,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15.
Submitted Solution:
```
from operator import itemgetter
class SegmentTree():
"""一点更新、区間取得クエリをそれぞれO(logN)で答えるデータ構造を構築する
update: i番目をvalに変更する
get_min: 区間[begin, end)の最小値を求める
"""
def __init__(self, n):
self.n = n
self.INF = 10**18
self.size = 1
while self.size < n:
self.size *= 2
self.node = [self.INF] * (2*self.size - 1)
def update(self, i, val):
i += (self.size - 1)
self.node[i] = val
while i > 0:
i = (i - 1) // 2
self.node[i] = min(self.node[2*i + 1], self.node[2*i + 2])
def get_min(self, begin, end):
begin += (self.size - 1)
end += (self.size - 1)
s = self.INF
while begin < end:
if (end - 1) & 1:
end -= 1
s = min(s, self.node[end])
if (begin - 1) & 1:
s = min(s, self.node[begin])
begin += 1
begin = (begin - 1) // 2
end = (end - 1) // 2
return s
n, k = map(int, input().split())
s = input()
info = []
for i, char in enumerate(s):
if char == "0":
info.append((i, i+1, i+1))
else:
info.append((max(i-k, 0), min(i+k+1, n), i+1))
info = sorted(info, key = itemgetter(1))
st = SegmentTree(n+1)
for begin, end, cost in info:
if begin == 0:
tmp = min(st.get_min(end - 1, end), cost)
else:
tmp = min(st.get_min(end - 1, end), st.get_min(begin - 1, end) + cost)
st.update(end - 1, tmp)
print(st.get_min(end - 1, end))
``` | instruction | 0 | 103,799 | 8 | 207,598 |
Yes | output | 1 | 103,799 | 8 | 207,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15.
Submitted Solution:
```
n, k = map(int, input().split())
s = input()
inf = 10**18
who = [-1] * n
for i in range(n):
if s[i] == '1':
for j in range(min(n - 1, i + k), i - 1, -1):
if who[j] == -1:
who[j] = i
else:
break
for i in range(n):
if s[i] == '1':
for j in range(i - 1, max(-1, i - k - 1), -1):
if who[j] == -1:
who[j] = i
else:
break
dp = [inf] * (n + 1)
dp[n] = 0
for i in range(n, 0, -1):
dp[i - 1] = min(dp[i - 1], dp[i] + i)
if who[i - 1] != -1:
dp[max(0, who[i - 1] - k)] = min(dp[max(0, who[i - 1] - k)], dp[i] + who[i - 1] + 1)
print(dp[0])
``` | instruction | 0 | 103,800 | 8 | 207,600 |
Yes | output | 1 | 103,800 | 8 | 207,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15.
Submitted Solution:
```
n,k=map(int,input().split())
s=list(input())
total=n*(n+1)//2
r=[i+1 for i in range(len(s)) if int(s[i])]
c=set()
for i in range(len(r)):
ma=max(1,r[i]-k)
mi=min(n,r[i]+k)
t=set(range(ma,mi+1))
if t.intersection(c)==t:
del r[i]
else:
c=c.union(t)
for i in r:
if i in s:
s.remove(i)
covered=sum(c)
router=sum(r)
print(total+router-covered)
``` | instruction | 0 | 103,801 | 8 | 207,602 |
No | output | 1 | 103,801 | 8 | 207,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15.
Submitted Solution:
```
n, k = map(int, input().split())
A = input()
# n = room, k = range of router
dp = [float("inf")] * (n)
for j in range(n):
if A[j] == '1':
dp[j] = min(dp[j], j + 1)
for l in range(k + 1):
if j + l < n:
dp[j + l] = min(dp[j + l], dp[j])
if j - l >= 0:
dp[j - l] = min(dp[j - l], dp[j])
ans = 0
directly = 0
for i, x in enumerate(dp):
if x > ans and x < float("inf"):
ans = x
if x == float("inf"):
directly += i + 1
print(ans + directly)
``` | instruction | 0 | 103,802 | 8 | 207,604 |
No | output | 1 | 103,802 | 8 | 207,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15.
Submitted Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify, nlargest
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()
##################
class Searchable_BIT():
def __init__(self,N):
self.N = N
self.node = [0]*(self.N+1)
self.cnt = 0
def add(self,a): # 要素 x を追加
x = a
self.cnt += 1
while x <= self.N:
self.node[x] += 1
x += x & -x
def delete(self,x): # 要素 x を削除
self.cnt -= 1
while x <= self.N:
self.node[x] -= 1
x += x & -x
def count(self,x): # x以下の要素数
if x==0: return 0
tmp = 0
while x > 0:
tmp += self.node[x]
x -= x & -x
return tmp
def get_maxval(self):
return self.get_lower_i(self.cnt)
def get_min_left(self,x): #x以上の最小の値を取得:
return self.get_lower_i(self.count(x-1)+1)
def get_lower_i(self,i): # i 番目に小さい要素を取得
NG = -1
OK = self.N
while OK-NG > 1:
mid = (OK+NG)//2
if self.count(mid) >= i:
OK = mid
else:
NG = mid
return OK
# se = [1,3,5,6,9]
# sbit = Searchable_BIT(100)
# for i in se: sbit.add(i)
# print(sbit.get_min_left(3)) #3
# print(sbit.get_min_left(4)) #5
# print(sbit.get_min_left(7)) #9
# print(sbit.get_min_left(20)) #100(max)
n,k = inpl()
s = [0] + inpsl(n)
sbit = Searchable_BIT(n+10)
for i,x in enumerate(s):
if x == '1':
sbit.add(i)
on = [0]*(n+1)
res = 0
# for i in range(1,n+1):
# print(i,sbit.get_min_left(i))
for i in range(1,n+1)[::-1]:
if on[i]: continue
mi = sbit.get_min_left(max(1,i-k))
if mi == n+10 or i+k < mi:
res += i
on[i] = 1
elif mi > i:
continue
else:
res += mi
sbit.delete(mi)
for j in range(max(1,mi-k),min(n+1,mi+k+1)):
on[j] = 1
off = [i for i in range(1,n+1) if on[i] == 0]
ofac = [0] + list(itertools.accumulate(off))
se = set()
for _,i in enumerate(off):
if i in se: continue
mi = sbit.get_min_left(max(1,i-k))
j = bisect_right(off,mi)
su = ofac[j]-ofac[_]
if mi < su:
res += mi
for k in range(_,j):
se.add(off[k])
else:
res += i
print(res)
``` | instruction | 0 | 103,803 | 8 | 207,606 |
No | output | 1 | 103,803 | 8 | 207,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.
You have to connect all n rooms to the Internet.
You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins.
Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers.
Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.
The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.
Output
Print one integer — the minimum total cost of connecting all n rooms to the Internet.
Examples
Input
5 2
00100
Output
3
Input
6 1
000000
Output
21
Input
4 1
0011
Output
4
Input
12 6
000010000100
Output
15
Note
In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.
In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.
In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.
In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15.
Submitted Solution:
```
a = input('').split(' ')
n = int(a[0])
k = int(a[1])
a = input('')
b = [0]*(n+1)
su = 0
i = 0
while(i<n):
if(a[i] == '1'):
su = su+i+1
e = min(n-1,i+k)
s = max(0,i-k)
b[e+1] = b[e+1]-1
b[s] = b[s]+1
if(e < n-1):
i = i+1
else:
break
else:
i = i+1
ans = [0]*(n+2)
for i in range(n+1):
ans[i+1] = ans[i]+b[i]
for i in range(1,n+1,1):
if(ans[i] == 0):
su =su+i
#print(ans)
print(su)
``` | instruction | 0 | 103,804 | 8 | 207,608 |
No | output | 1 | 103,804 | 8 | 207,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1]. | instruction | 0 | 103,887 | 8 | 207,774 |
Tags: binary search, greedy, math, sortings
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
l = list(map(int,input().split()))
a = 0
s = sum(l)
if s%(n-1)==0:
m = max(l)
if m>s//(n-1):
print(a+(m-s//(n-1))*(n-1))
else:
print(a)
else:
a+=(n-1)-s%(n-1)
s+=a
m = max(l)
if m>s//(n-1):
print(a+(m-s//(n-1))*(n-1))
else:
print(a)
``` | output | 1 | 103,887 | 8 | 207,775 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1]. | instruction | 0 | 103,888 | 8 | 207,776 |
Tags: binary search, greedy, math, sortings
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import math
def main():
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
print(max(max(a),math.ceil(float(sum(a))/(n-1)))*(n-1)-sum(a))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 103,888 | 8 | 207,777 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1]. | instruction | 0 | 103,889 | 8 | 207,778 |
Tags: binary search, greedy, math, sortings
Correct Solution:
```
p=int(input())
for i in range(p):
n=int(input())
minn=0
maxx=0
sm=0
k=list(map(int,input().split()))
for i in range(n):
if i==0:
minn=k[i]
maxx=k[i]
else:
if minn>k[i]:
minn=k[i]
if maxx<k[i]:
maxx=k[i]
sm+=k[i]
r=maxx*(n-1)-sm+minn
if minn>r:
if n-1==1:
print(0)
else:
if (minn-r)%(n-1)==0:
print(0)
else:
otv=(n-1)-(minn-r)%(n-1)
print(otv)
else:
print(r-minn)
``` | output | 1 | 103,889 | 8 | 207,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1]. | instruction | 0 | 103,890 | 8 | 207,780 |
Tags: binary search, greedy, math, sortings
Correct Solution:
```
import math
t = int(input())
while t:
n = int(input())
a = list(map(int,input().split()))
total = sum(a)
k = max(math.ceil(total/(n-1)), max(a))
ans = (n-1)*k - total
print(ans)
t-=1
``` | output | 1 | 103,890 | 8 | 207,781 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1]. | instruction | 0 | 103,891 | 8 | 207,782 |
Tags: binary search, greedy, math, sortings
Correct Solution:
```
# cook your dish here
import math
t = int(input())
for i in range(t):
n = int(input())
a = list(map(int,input().split()))
s,m = sum(a),max(a)
ans = max(m,math.ceil(s/(n-1)))*(n-1)-s
print(ans)
``` | output | 1 | 103,891 | 8 | 207,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1]. | instruction | 0 | 103,892 | 8 | 207,784 |
Tags: binary search, greedy, math, sortings
Correct Solution:
```
import sys
import math
from collections import Counter
from collections import OrderedDict
from collections import defaultdict
from functools import reduce
#from itertools import groupby
sys.setrecursionlimit(10**6)
def inputt():
return sys.stdin.readline().strip()
def printt(n):
sys.stdout.write(str(n)+'\n')
def listt():
return [int(i) for i in inputt().split()]
def gcd(a,b):
return math.gcd(a,b)
def lcm(a,b):
return (a*b) // gcd(a,b)
def factors(n):
step = 2 if n%2 else 1
return set(reduce(list.__add__,([i, n//i] for i in range(1, int(math.sqrt(n))+1, step) if n % i == 0)))
def comb(n,k):
factn=math.factorial(n)
factk=math.factorial(k)
fact=math.factorial(n-k)
ans=factn//(factk*fact)
return ans
def is_prime(n):
if n <= 1:
return False
if n == 2:
return True
if n > 2 and n % 2 == 0:
return False
max_div = math.floor(math.sqrt(n))
for i in range(3, 1 + max_div, 2):
if n % i == 0:
return False
return True
def maxpower(n,x):
B_max = int(math.log(n, x)) + 1 #tells upto what power of x n is less than it like 1024->5^4
return B_max
t=int(inputt())
#t=1
for _ in range(t):
n=int(inputt())
l=listt()
maxi=max(l)
s=sum(l)
#k=max(maxi,math.ceil(s//(n-1)))
#ans=(k*(n-1))-s
#print(abs(ans))
print(max(maxi,math.ceil(s/(n-1)))*(n-1)-s)
``` | output | 1 | 103,892 | 8 | 207,785 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1]. | instruction | 0 | 103,893 | 8 | 207,786 |
Tags: binary search, greedy, math, sortings
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int,input().split()))
x = max(a)*(n-1)
s = sum(a)
q = s//(n-1)
r = s%(n-1)
if r>0:
q+=1
print(max(max(a),q)*(n-1)-s)
``` | output | 1 | 103,893 | 8 | 207,787 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1]. | instruction | 0 | 103,894 | 8 | 207,788 |
Tags: binary search, greedy, math, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import *
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
M = (n-1)*max(a)
S = sum(a)
if S<=M:
#print('a')
print(M-S)
else:
tot = M+(S-M+(n-2))//(n-1)*(n-1)
print(tot-S)
``` | output | 1 | 103,894 | 8 | 207,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1].
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
A = list(map(int, input().split()))
A.sort(reverse=True)
ans = 0
maxv = A[0]
for i in range(1, n - 1):
ans += maxv - A[i]
ans = max(0, ans - A[-1])
r = (ans + sum(A)) % (n - 1)
if not r:
print(ans)
else:
ans += n - 1 - r
print(ans)
``` | instruction | 0 | 103,895 | 8 | 207,790 |
Yes | output | 1 | 103,895 | 8 | 207,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1].
Submitted Solution:
```
import math
from sys import stdin,stdout
mp=lambda:map(int,stdin.readline().split())
li=lambda:list(map(int,stdin.readline().split()))
for _ in range(int(input())):
n=int(input())
c=li()
mx1=max(c)
sm=sum(c)
mx2=(sm-1)//(n-1) + 1
mx=max(mx1,mx2)
print((n-1)*mx-sm)
``` | instruction | 0 | 103,896 | 8 | 207,792 |
Yes | output | 1 | 103,896 | 8 | 207,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are asked to watch your nephew who likes to play with toy blocks in a strange way.
He has n boxes and the i-th box has a_i blocks. His game consists of two steps:
1. he chooses an arbitrary box i;
2. he tries to move all blocks from the i-th box to other boxes.
If he can make the same number of blocks in each of n - 1 other boxes then he will be happy, otherwise, will be sad. Note that your nephew can only move the blocks from the chosen box to the other boxes; he cannot move blocks from the other boxes.
You don't want to make your nephew sad, so you decided to put several extra blocks into some boxes in such a way that no matter which box i he chooses he won't be sad. What is the minimum number of extra blocks you need to put?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (2 ≤ n ≤ 10^5) — the number of boxes.
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the number of blocks in each box.
It's guaranteed that the sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print a single integer — the minimum number of blocks you need to put. It can be proved that the answer always exists, i. e. the number of blocks is finite.
Example
Input
3
3
3 2 2
4
2 2 3 2
3
0 3 0
Output
1
0
3
Note
In the first test case, you can, for example, put one extra block into the first box and make a = [4, 2, 2]. If your nephew chooses the box with 4 blocks, then we will move two blocks to the second box and two blocks to the third box. If he chooses the box with 2 blocks then he will move these two blocks to the other box with 2 blocks.
In the second test case, you don't need to put any extra blocks, since no matter which box your nephew chooses, he can always make other boxes equal.
In the third test case, you should put 3 extra blocks. For example, you can put 2 blocks in the first box and 1 block in the third box. You'll get array a = [2, 3, 1].
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
nums = list(map(int, input().split()))
s = sum(nums)
ma = max(nums)
lim, r = divmod(s, n-1)
if r:
lim += 1
print(max(lim, ma)*(n-1)-s)
``` | instruction | 0 | 103,897 | 8 | 207,794 |
Yes | output | 1 | 103,897 | 8 | 207,795 |
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