**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++ :
/*
*/
#pragma comment(linker, "/STACK:16777216")
#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 yn askfhwqriuperikldjk
#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;
}
bool good_mask(int mask)
{
int ar[25];
for (int i = 0; i < n; i++)
{
ar[i] = 1;
}
for (int i = 0; i < n; i++)
{
if (mask&(1 << 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 ans_mask;
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;
/*for (int mask = 0; mask < (1 << n); mask++)
{
if (good_mask(mask))
{
int here = 0;
for (int i = 0; i < n; i++)
{
if (mask&(1 << i))
here += ar[i + 1];
}
if (here < ans)
ans = min(ans, 0ll + here),
ans_mask = mask;
}
}*/
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);
}
}
```

## View More Similar Problems

## Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

View Solution →## Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

View Solution →## Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

View Solution →## Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

View Solution →## Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

View Solution →## Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty

View Solution →