UVA11332

Problem J: Summing Digits

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.
Each line of input contains a single positive integer n at most 2,000,000,000. For each such integer, you are to output a single line containing g(n). Input is terminated by n = 0 which should not be processed.

Sample input

2
11
47
1234567892
0

Output for sample input

2
2
2
2

import java.util.*;

public class UVA11332 {

 public static void main(String[] args) {

  Scanner sc = new Scanner(System.in);

  while (sc.hasNextInt()) {

   long n = sc.nextInt();

   if (n == 0)
    break;

   while (n / 10 > 0) {

    int sum = 0;

    while (n > 0) {
     sum += n % 10;

     n /= 10;
    }

    n = sum;
   }
   System.out.println(n);
  }

 }

}