Jarvis and Seven Segments

problem:-

All over the world, peoples are working on energy solution. It would be a tough time for our next generation to survive if we don’t think about solution. Tony stark is working on a new project and wants to display his project using “seven segment display - concept”. Tony Stark gave Jarvis a task to find a number from his Favorite list of number for which the energy consumption is lowest.
(Assuming that for a digit to represent Tony stark is using 7 bulbs and only those bulbs light up which are required to represent a number and rest other would be completely off.)
Help Jarvis and conserve energy.
Seven segment display - https://en.wikipedia.org/wiki/Seven-segment_display

Input:
First line will contain the number of test cases and for every test case first line will contain length of favorite list and the second line for a test case will contain n numbers
Output:
For every test case print the answer. If there exist more than 1 numbers for which same number of bulbs are required than output the number which occurs first in the Favorite list.
Constraints:
Test cases< 10
A[i] < 10^6
Size of list < 10^5

SAMPLE INPUT

 

1
5
1 2 3 4 5

SAMPLE OUTPUT

 

1

Explanation

Number 1 needs only two bulbs to represent.
Solution:-

#include<stdio.h>

int main()

{

  long int t,n,r,sum=0,ans,flag=0,i,j;

  long int a,new_num,n1;

  scanf("%ld",&t);

  for(i=0;i<t;i++)

  {

    flag=0;

    scanf("%ld",&n);

    for( j=0;j<n;j++)

    {

     scanf("%ld",&a);

       n1=a;

       if(a==0)

       sum=sum+6;

     while(a!=0)

     {

r=a%10;

       if(r==1)

       sum=sum+2;

       if(r==2||r==3||r==5)

       sum=sum+5;

         if(r==4)

         sum=sum+4;

         if(r==6||r==9||r==0)

         sum=sum+6;

         if(r==7)

         sum=sum+3;

         if(r==8)

         sum=sum+7;

         a=a/10;

     }

      

if(flag==0)

       {

         ans=sum;

         new_num=n1;

         flag=1;

       }

       if(ans>sum)

       {

       ans=sum;

        new_num=n1;

       }

       sum=0;

    }

    printf("%ld\n",new_num);

  }

  return 0;

}