SORT17

QUESTION DESCRIPTION 

Given two arrays, A and B, of equal size n, the task is to find the minimum value of A[0] * B[0] + A[1] * B[1] +…+ A[n-1] * B[n-1], where shuffling of elements of arrays A and B is allowed.

Madatory conditions are "void result(int a[],int b[],int n)"
Examples:

Input : A[] = {3, 1, 1} and B[] = {6, 5, 4}.
Output : 23 Minimum value of S = 1*6 + 1*5 + 3*4 = 23.

Input : A[] = { 6, 1, 9, 5, 4 } and B[] = { 3, 4, 8, 2, 4 }
Output : 80. Minimum value of S = 1*8 + 4*4 + 5*4 + 6*3 + 9*2 = 80.


Input:
The first line of input contains an integer denoting the no of test cases.

Then T test cases follow. Each test case contains three lines.

The first line of input contains an integer N denoting the size of the arrays.

In the second line are N space separated values of the array A[], and

In the last line are N space separated values of the array B[].

Output:

For each test case in a new line print the required result.

Constraints:

1<=T<=100

1<=N<=50

1<=A[]<=20

TEST CASE 1 

INPUT

2
3 
3 1 1
6 5 4
5
6 1 9 5 4
3 4 8 2 4

OUTPUT

23
80

TEST CASE 2 

INPUT

2
4
3 8 1 6
1 7 5 1
5
6 4 7 1 3
4 2 6 9 1

OUTPUT

36
62

#include <stdio.h>
void result(int a[],int b[],int n);
int main()
{
  int t, i, j, arr1[30], arr2[30], x, temp;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%d",&x);
    for(i=0;i<x;i++)
    {
      scanf("%d",&arr1[i]);
    }
   
    for(i=0;i<x;++i)
    {
     for (j=i+1;j<x;++j)
      {
       if (arr1[i] > arr1[j])
        {
         temp = arr1[i];
         arr1[i] = arr1[j];
         arr1[j] = temp;
        }
      }
    }
   
    for(i=0;i<x;i++)
    {
      scanf("%d",&arr2[i]);
    }
   
    for(i=0;i<x;++i)
    {
     for (j=i+1;j<x;++j)
      {
       if (arr2[i] < arr2[j])
        {
         temp = arr2[i];
         arr2[i] = arr2[j];
         arr2[j] = temp;
        }
      }
    }
   
    result(arr1, arr2, x);
  }
  return 0;
}

void result(int a[],int b[],int n)
{
  int i, fin=0, sum;
  for(i=0;i<n;i++)
  {
      fin+=a[i]*b[i];
      sum=fin;
  }
  printf("%d\n",sum);
}