Solution of 11332 - Summing Digits

Problem Description
source:https://uva.onlinejudge.org/external/113/11332.html

For a positive integer n, let f(n) denote the sum of the digits of n when represented in base 10. It is easy to see that the sequence of numbers n, f(n), f(f(n)), f(f(f(n))), . . . eventually becomes a single digit number that repeats forever. Let this single digit be denoted g(n). 
For example, consider n = 1234567892. Then: 

    f(n) = 1+2+3+4+5+6+7+8+9+2 = 47 
    f(f(n)) = 4 + 7 = 11 
    f(f(f(n))) = 1 + 1 = 2 
    Therefore, g(1234567892) = 2. 

Input 

Each line of input contains a single positive integer n at most 2,000,000,000. Input is terminated by n = 0 which should not be processed. 

Output 

For each such integer, you are to output a single line containing g(n). 

Sample Input 

2 
11 
47 
1234567892 
0 

Sample Output 

2 
2 
2 
2

Solution:

#include<stdio.h>
int SumDgit(long int);
int main()
{
     long int n,value;
    while(scanf("%ld",&n)==1)
    {
            if(n==0)
                break;
           SumDgit(n);
    }
    return 0;
}

int SumDgit(long int n)
{
    long int sum,get;
    sum=0;
    while(n!=0)
    {
        sum=sum+n%10;
        n=n/10;
        get=sum;
    }
    if(get/10==0)
        printf("%ld\n",get);
    else
        SumDgit(get);

}