Solution of 10018 - Reverse and Add

Problem Description
source:https://uva.onlinejudge.org/external/100/10018.html

The “reverse and add” method is simple: choose a number, reverse its digits and add it to the original. If the sum is not a palindrome (which means, it is not the same number from left to right and right to left), repeat this procedure.

For example: 
                       195 Initial number 
                       591 
                       —– 
                       786 
                       687 
                      —– 
                     1473 
                     3741 
                     —– 
                    5214 
                    4125 
                    —– 
                    9339   Resulting palindrome


     In this particular case the palindrome ‘9339’ appeared after the 4th addition. This method leads to palindromes in a few step for almost all of the integers. But there are interesting exceptions. 196 is the first number for which no palindrome has been found. It is not proven though, that there is no such a palindrome.
    You must write a program that give the resulting palindrome and the number of iterations (additions) to compute the palindrome.
    You might assume that all tests data on this problem:
        • will have an answer ,
        • will be computable with less than 1000 iterations (additions),
        • will yield a palindrome that is not greater than 4,294,967,295.

Input 

The first line will have a number N (0 < N ≤ 100) with the number of test cases, the next N lines will have a number P to compute its palindrome.

Output

For each of the N tests you will have to write a line with the following data : minimumnumberofiterations(additions)togettothepalindrome and theresultingpalindromeitself separated by one space.

Sample Input 

3 
195 
265 
750 

Sample Output 

4 9339 
5 45254 
3 6666

Solution: 

#include<stdio.h>
long long int Reverse(long long int n);
int main()
{
    long long int t, n,rev,sum,sumr, i,j,step, number, number1;
    while(scanf("%lld",&t)==1)
    {
        for(i=1;i<=t;i++)
        {
            scanf("%lld",&n);
            step=0;
            number1=n;
            number=Reverse(n);
            if(number==number1)
              printf("0 %lld\n",number);
            else
            {
                do
                {
                    sum=number+number1;
                    number=sum;
                    number1=Reverse(sum);
                    step++;

                }while(number!=number1);
                printf("%lld %lld\n",step,number);
            }
        }
    }
    return 0;
}

long long int Reverse(long long int n)
{
    long long int r;
    r=0;
    while(n!=0)
    {
        r=r*10+(n%10);
        n=n/10;
    }
    return r;
}