Even Length Palindromic Number

Problem:-

You have to design a new model which maps an even length palindromic number to some digit between 0 to 9.
The number is mapped to a digit $x$ on the basis of following criteria:
1. $x$ should appear maximum number of times in the palindromic number, that is, among all digits in the number, $x$ should appear maximum number of times.
2. If more than one digit appears maximum number of times, $x$ should be the smallest digit among them.

Given an integer $N$, you have to find the digit $x$ for the $N^{th}$ even length palindromic number.
Note- First 9 even length palindromic numbers are:
            11, 22, 33, 44, 55, 66, 77, 88, 99
Input :
    First line contains T, number of test cases.
    Each of the next T lines contains an integer N.
Output:
    For each test case, print the digit to which the $N^{th}$ even length palindromic number is mapped.
    Answer for each test case should come in a new line.
Constraints:
    $1\le T\le {10}^5$
    $1\le N\le {10}^{18}$

SAMPLE INPUT

 

3
1
2
10

SAMPLE OUTPUT

 

1
2
0

Explanation

For case 1:
    1st even length palidromic number is 11 , so answer is 1 as 1 appears most number of times in the number.
For case 2:
    2nd even length palidromic number is 22 , so answer is 2 as 2 appears most number of times in the number.
For case 3:
    10th even length palindromic number is 1001, here both 0 and 1 appears same number of times but 0<1 so answer is 0.

Time Limit:1.0 sec(s) for each input file.

Memory Limit:256 MB

Source Limit:1024 KB

solution:-

#include<stdio.h>

void main()

{

     int t,k=0,a[10],max,r;

     long long int n;

     scanf("%d",&t);

     for(int i=0;i<t;i++)

     {

        for(int j=0;j<=9;j++)

         a[j]=0;

        scanf("%lld",&n);

        while(n!=0)

        {

         r=n%10;

         a[r]++;

         n=n/10;

        }

        max=a[0];

        for(int j=0;j<=9;j++)

        {

if(max<a[j])

            {

             max=a[j];

k=j;

            }

        }

        printf("%d\n",k);

        k=0;

     }

}