UVa 11332 Summing Digits (water ver.)

11332 - Summing Digits

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2307

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.

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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

完整代码：

/*0.016s*/

#include<cstdio>
#include<cstring>  

char n[15];  

int main()
{
    int i, sum, len;
    while (gets(n), n[0] != '0')
    {
        sum = 0;
        len = strlen(n);
        for (i = 0; i < len; ++i)
            sum += n[i] & 15;
        sum = sum / 10 + sum % 10;
        sum = sum / 10 + sum % 10;
        printf("%d\n", sum);
    }
    return 0;
}

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