UVa 11332 Summing Digits （water ver.）

UVa 11332 Summing Digits （water ver.）2014-10-06 csdn博客 synapse711332 - 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
", sum）;}return 0;}