Even length Palindromic Number

program:-

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 

Nth

 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 

Nth

 even length palindromic number is mapped.
    Answer for each test case should come in a new line.
Constraints:
    

1T105


    

1N1018

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

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(“%dn”,k);
}


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