Minimum sum

Given an array of digits (values are from 0 to 9), find the minimumpossible sum of two numbers formed from digits of the array. All digits of given array must be used to form the two numbers.

Input:

The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.

Output:

Corresponding to each test case, in a new line, print the minimum possible sum of two numbers formed from digits of the array.

Constraints:

1 ≤ T ≤ 100
1 ≤ N ≤ 30
1 ≤ A[i] ≤ 9

Example:

Input
1
5
5 3 0 7 4

Output
82

Your Code

#include<iostream>

using namespace std;

#include<math.h>

int main()

{

int t;

cin>>t;

while(t--)

{

int a[30],n;

cin>>n;

int i;

for(i=0; i<n; i++)

{

cin>>a[i];

}

int j;

for(i=0; i<n-1; i++)

{    int c=i;

for(j=i+1; j<n; j++)

{ 

if(a[c]>a[j])

{

c=j;

}

}

if(c!=i)

{

int temp=a[i];

a[i]=a[c];

a[c]=temp;

}

} 

long long int num1=0,num2=0;

int k=0;

for(i=n-1; i>=0; i--)

{    // cout<<a[i]<<endl;

num1=num1+a[i]*pow(10,k);

i--;

if(i>=0)

{

num2=num2+a[i]*pow(10,k);

     

//cout<<a[i]<<endl;

     }

k++;

//cout<<num1<<" "<<num2;

}

cout<<num1+num2<<endl;

}

}