# Turn Off the Lights

### Problem Statement :

```There are n bulbs in a straight line, numbered from 0 to n-1. Each bulb i has a button associated with it, and there is a cost,ci , for pressing this button. When some button i is pressed, all the bulbs at a distance <= k from bulb  will be toggled(off->on, on->off).

Given n, k, and the costs for each button, find and print the minimum cost of turning off all n bulbs if they're all on initially.

Input Format

The first line contains two space-separated integers describing the respective values of n and k.
The second line contains n space-separated integers describing the respective costs of each bulb (i.e., c0,c1,c2,...,cn-1).

Constraints
3 <= n <= 10^4
0 <= k <= 1000
0 <= ci <= 10^9
Output Format

Print a long integer denoting the minimum cost of turning off all n bulbs.```

### Solution :

```                            ```Solution in C :

In C++ :

/*
*/

#define _CRT_SECURE_NO_WARNINGS

#include <fstream>
#include <iostream>
#include <string>
#include <complex>
#include <math.h>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <memory.h>
#include <assert.h>

#define y0 sdkfaslhagaklsldk
#define y1 aasdfasdfasdf
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash
#define norm asdfasdgasdgsd

#define eps 1e-9
#define M_PI 3.141592653589793
#define bs 1000000007
#define bsize 350

using namespace std;

const int INF = 1e9;
const int N = 200031;

int n, k, ar[N];

long long solve(int start)
{
long long res = 0;
for (int i = start; i <= n; i += 2 * k + 1)
{
res += ar[i];
if (i + k + 1 <= n&&i + k * 2 + 1 > n)
return 1e18;
}
return res;
}

{
int ar[25];

for (int i = 0; i < n; i++)
{
ar[i] = 1;
}
for (int i = 0; i < n; i++)
{
{
for (int j = i - k; j <= i + k; j++)
{
if (j >= 0 && j < n)
ar[j] ^= 1;
}
}
}
for (int i = 0; i < n; i++)
{
if (ar[i])
return 0;
}
return 1;
}

int main(){
//freopen("fabro.in","r",stdin);
//freopen("fabro.out","w",stdout);
//freopen("F:/in.txt", "r", stdin);
//freopen("F:/output.txt", "w", stdout);
ios_base::sync_with_stdio(0);
//cin.tie(0);

while (true)
{
cin >> n >> k;
for (int i = 1; i <= n; i++)
{
cin >> ar[i];
//ar[i] = rand() % 10;
}

long long ans = 1e18;

{
{
int here = 0;
for (int i = 0; i < n; i++)
{
here += ar[i + 1];
}
if (here < ans)
ans = min(ans, 0ll + here),
}
}*/
long long ans2 = 1e18;

for (int F = 1; F <= k + 1&&F<=n; F++)
{
ans2 = min(ans2, solve(F));
}

/*if (ans != ans2)
{
cout << "!" << endl;
for (int i = 1; i <= n; i++)
{
cout << ar[i] << " ";
}
cout << endl;
cout << ans << " " << ans2 << " " << ans_mask << endl;
while (true);
}

else
cout << "OK" << endl;*/

cout << ans2 << endl;
return 0;
}
cin.get(); cin.get();
return 0;
}

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
int[] c = new int[n];
for (int i = 0; i < n; i++) {
c[i] = sc.nextInt();
}
long ans = 1000000000000000000l;
for (int i = 0; i <= Math.min(n-1, k); i++) {
int maxCovered = i+k;
long partialAns = c[i];
boolean possible = true;
while (maxCovered < n-1) {
int button = maxCovered+k+1;
if (button >= n) {
possible = false;
break;
}
partialAns += c[button];
maxCovered = button+k;
}
if (!possible)
continue;
ans = Math.min(partialAns, ans);
}
System.out.println(ans);
}
}```
```

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