Maximise the Minimum solution|Codesheff starter 23 for div 3 (Rated)

You are given two arrays A$A$ and B$B$, both of length N$N$.

You would like to choose exactly K$K$ distinct indices i1,i2,…,iK$i_1,i_2,\dots ,i_K$ such that min(Ai1+Ai2+…+AiK,Bi1+Bi2+…+BiK)$min(A_{i_1}+A_{i_2}+\dots +A_{i_K},B_{i_1}+B_{i_2}+\dots +B_{i_K})$ is maximized. Find this maximum value.

Input Format

• The first line of input contains a single integer T$T$, denoting the number of test cases. The description of T$T$ test cases follows.
• Each test case consists of 3$3$ lines of input.
  • The first line of each test case contains two space-separated integers N$N$ and K$K$.
  • The second line contains N$N$ space-separated integers A1,A2,…,AN$A_1,A_2,\dots ,A_N$ denoting the elements of A$A$.
  • The third line contains N$N$ space-separated integers B1,B2,…,BN$B_1,B_2,\dots ,B_N$ denoting the elements of B$B$.

Output Format

For each test case, output a new line containing one integer — the required maximum value.

Constraints

• 1≤T≤20$1\le T\le 20$
• 2≤N≤40$2\le N\le 40$
• 1≤K≤N$1\le K\le N$
• 1≤Ai,Bi≤40$1\le A_i,B_i\le 40$

Sample Input 1 

3
5 3
4 2 3 1 4
3 2 5 5 1
4 2
1 2 3 4
4 3 2 1
6 3
8 10 3 6 7 2
4 8 4 1 1 6

Sample Output 1 

9
5
18

Explanation

Test Case 1$1$: One optimal choice of indices is i1=1,i2=3,i3=5$i_1=1,i_2=3,i_3=5$. This gives A1+A3+A5=11$A_1+A_3+A_5=11$ and B1+B3+B5=9$B_1+B_3+B_5=9$.

Test Case 2$2$: One optimal choice of indices is i1=1,i2=4$i_1=1,i_2=4$.

Test Case 3$3$: One optimal choice of indices is i1=1,i2=2,i3=6$i_1=1,i_2=2,i_3=6$.

Solution:

#include <iostream>

using namespace std;

int main()

{

   int t;

   cin>>t;

   while(t--){

       int n,k,max=0;

       cin>>n>>k;

       int a[n],b[n];

       for(int i=1;i<=n;i++){

           cin>>a[i];         

       }

       for(int i=1;i<=n;i++){

           cin>>b[i];

       }

       int sum1=0,sum2=0;

       for(int i=1;i<=n;i++){

           for(int j=1;j<=n;j++){

               max=(a[i],b[i]);

               sum1+=a[i];

               sum2+=b[i];

           }  

       }

       if(sum1>sum2){

           cout<<sum1<<endl;

       }else{

           cout<<sum2<<endl;

       }

   }

    return 0;

}