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
enter image description here
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(“%ldn”,new_num);
  }
  return 0;
}

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