1955-digit root

【C系列3.11】中文简单digit root 1955

Time Limit:  1
s      Memory Limit:   32
MB

Submission：357    
AC：175    
Score：10.00


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Status

Description

一个正数的digit root就是这个正数每个数字（个位数字，十位数字等）的和，如果这个和包含两个或更多数字，这个过程就要重复，直到最终得到一个数。（简单化自hdoj1013）如果本人翻译太烂。。。下面是原文。。。

The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains
two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.

For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the
3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.

Input

每行输入一个数字n（1 <= n <= 100000）。

Output

输出n的digit root.

Samples

input:

12398561

output:

62

下附AC代码：

#include<stdio.h>
#include<string.h>
int f(int n) {
int s = 0;
while (n) {
s += n % 10;
n /= 10;
}
return s;
}
int main() {
int i;
char str[1000];
while (gets(str)) {
int n = 0;
if (str[0] == '0')
break;
for (i = 0; i < strlen(str); ++i)
n += str[i] - '0';
while (n > 9)
n = f(n);
printf("%d\n", f(n));
}
}

原题链接：http://acm.hznu.edu.cn/OJ/problem.php?cid=1091&pid=28