HDU - 1310 - Digital Roots

先上题目：

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39890    Accepted Submission(s): 12286

Problem Description

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

The
input file will contain a list of positive integers, one per line. The
end of the input will be indicated by an integer value of zero.

 

Output

For each integer in the input, output its digital root on a separate line of the output.

 

Sample Input

24
39
0

 

Sample Output

6
3

 

　　题意是给你一个数字，求出它的各位数字之和，如果结果不是小于10的，就继续前面的操作，输出最后结果。

　　其实这一题没有什么好说，就一个地方有点坑，明明说好了是integer，输入竟然有可能超过int，要用字符串读入才可以过= =。

 

上代码：

 

#include <cstdio>
#include <cstring>

using namespace std;

char s[];

int main()
{
    int n,a,len;
    //freopen("data.txt","r",stdin);
    while()
    {
        scanf("%s",s);
        n=;
        len=strlen(s);
        while(len--){n+=s[len]-'';}
        if(n==) break;
        while(n>=)
        {
            a=;
            while(n){a+=n%; n/=;}
            n=a;
        }

        printf("%d\n",n);
    }
    return ;
}

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