MEGA SALE

PROBLEM :

LALU wanted to purchase a laptop so he went to a nearby sale.There were n Laptops at a sale. Laptop with index i costs ai rupees. Some Laptops have a negative price — their owners are ready to pay LALU if he buys their useless Laptop. LALU can buy any Laptop he wants. Though he's very strong, he can carry at most m Laptops, and he has no desire to go to the sale for the second time. Please, help LALU find out the maximum sum of money that he can earn.

Input:
First line of the input contains T denoting the number of test cases.Each test case has 2 lines :

first line has two spaced integers n m.
second line has n integers [a0...ai...an-1].

Output:
The maximum sum of money that LALU can earn, given that he can carry at most m Laptops.


Constraints:
1=T=10
1=n,m=100
-1000=ai=1000


Sample Input:
1
5 3
-6 0 35 -2 4

Sample Output:
8

Explanation:
LALU takes the laptops with -6 and -2 and thus earns 8 rupees.

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SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------

#include<iostream>
using namespace std;
int MEGA_SALE(int [],int ,int ) ;
void bubble_sort(int [],int ) ;
int minof(int ,int ) ;
int main()
 {
int t,arr[100],no,i,k ;
cin>>t ;
while(t--)
{
   cin>>no ;
   cin>>k ;
   for(i=0;i<no;i++)
       cin>>arr[i] ;
     
   no=MEGA_SALE(arr,no,k) ;
   cout<<abs(no)<<endl ;
}
return 0;
}

int MEGA_SALE(int arr[],int no,int k)
{
    int i ;
    bubble_sort(arr,no) ;
   
    int sum=0 ;
    for(i=0;i<k;i++)
        sum=minof(sum,sum+arr[i]) ;
       
    return sum ;
}

void bubble_sort(int arr[],int no)
{
    int i,j,temp ;
    for(i=0;i<no-1;i++)
    {
        for(j=0;j<no-i-1;j++)
        {
            if(arr[j]>arr[j+1])
            {
                temp=arr[j] ;
                arr[j]=arr[j+1] ;
                arr[j+1]=temp ;
            }
        }
    }
}

int minof(int a,int b)
{
    return a>b?b:a ;
}      

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