2011-12-14 23:04:00Morris
[UVA] 11332 - Summing Digits
Problem J: Summing Digits
For a positive integer n, let f(n) denote the sum of the digitsof n when represented in base 10. It is easy to see that the sequence ofnumbers n, f(n), f(f(n)), f(f(f(n))), ... eventually becomes a single digitnumber 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 = 47f(f(n)) = 4+7 = 11f(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
2114712345678920
Output for sample input
2222
#include<stdio.h>
int f(int n) {
if(n < 10) return n;
int sum = 0;
while(n) {
sum += n%10;
n /= 10;
}
return f(sum);
}
int main() {
int n;
while(scanf("%d", &n) == 1 && n)
printf("%d\n", f(n));
return 0;
}