Min_No.of_Laddus.cpp

// Chhota Bheem is fond of Laddus, there are N boxes in a row, each box contains 
// some laddus, where k-th box conatins box[k]-number of laddus.

// In one go, Choota Bheem can choose any box and eat floor(box[i]/2) laddus 
// from it. He has do the above step, exactly K times. He is allowed to eat 
// from the same box any number of times.

 
// You are given an list of N integers, number of laddus in each box.
// Your task is to return the minimum number of laddus left in all the boxes 
// together, after K times.

// NOTE: floor(I) is the greatest integer that is smaller than or equal to I.
 
// Input Format:
// -------------
// Line -1: Two space separated integers N and K,number of boxes and value of K.
// Line -2: N space separated integers, number of laddus in each box.
 
// Output Format:
// --------------
// Print an integer result.
 
// Sample Input-1:
// ---------------
// 3 2
// 3 5 9
 
// Sample Output-1:
// ----------------
// 11
 
// Explanation:
// ------------
// Steps of a possible scenario are:
// - Chota bheem selected box-2. The resulting boxes are [3,5,5].
// - Chota bheem selected box-2. The resulting boxes are [3,5,3].
// The total number of laddus in all the boxes [3,5,3] is 11.

 
// Sample Input-2:
// ---------------
// 5 3
// 4 3 8 6 7
 
// Sample Output-2:
// ----------------
// 18



Solution 1:   //using Priority_Queue
#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,k;
    cin>>n>>k;
    priority_queue<int> q;
    for(int i=0;i<n;i++){
        int x;
        cin>>x;
        q.push(x);
    }
    while(k>0){
        int x=q.top();
        q.pop();
        if(x%2!=0){
            q.push(x/2+1);
        }
        else{
            q.push(x/2);
        }
        k--;
        
    }
    int sum=0;
    while(!q.empty()){
        sum=sum+q.top();
        q.pop();
    }
    cout<<sum;
}

......................................................

Solution 2: //Using Vector
#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,k;
    cin>>n>>k;
    vector<int> v(n,0);
    for(int i=0;i<n;i++){
        cin>>v[i];
    }
    while(k>0){
        sort(v.begin(),v.end());
        int x=v[v.size()-1];
        if(x%2!=0){
            v[v.size()-1]=x/2+1;
        }
        else{
            v[v.size()-1]=x/2;
        }
        k--;
    }
    int sum=0;
    for(auto i:v){
        sum=sum+i;
    }
    cout<<sum;
}