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

#include<stdio.h>
int main()
{
    int n,sum=0;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        sum=0;
        while(n!=0)
        {
            sum+=n%10;
            n/=10;
            if(sum>10&&n==0)
            {    
                n=sum;
                sum=0;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}

我这个程序为什么通过不了的，，显示Wrong answer

你应该考虑一下sum的值 这个应该要用到递归!我再你得题目上面改一下!

#include<stdio.h>
int sum;
void f(int n){
    sum+=n%10;
    n/=10;
    if(n>9)
        f(n);
    else
        sum+=n;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        sum=0;
        f(n);
        printf("%d\n",sum);
        while(sum>9)
        {
            int k=sum;
            sum=0;
            f(k);
        }
        printf("%d\n",sum);
    }
    return 0;
}

互相探讨一下我觉得应该把 sum放进 main 里面~ 而作为函数参数传递 ~ 以减少全局函数~
还有main里面只调用一次f()；就能直接输出结果 ~

#include<stdio.h>
int f(int n,int sum){
    sum+=n%10;
    n/=10;
    if(n>9)
        sum=f(n,sum);
    else
        sum+=n;
    if(sum>9)
    {
        int k=sum;
        sum=0;
        sum+=f(k,sum);
    }
    return sum;
}
int main()
{
    int n,sum;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        sum=0;
        sum=f(n,sum);
        printf("%d\n",sum);
    }
    return 0;
}

思考了下发现有意思哦~ 题目没要求负数怎么办！ 还有没说 要转换的数有没有界限！~ 如果超过了界限 咱就不能用int来表示了！！！

AC代码

#include <stdio.h>
#include <string.h>
char a[1000];
int main()           
{
    int i,j,k;
    char temp[10];
    while(scanf("%s",a) && a[0]-48)
    {
        int len = strlen(a);
        int sum = 0;
        for(i = 0;a[i];i++)sum+=a[i]-48;
        while(sum>9)
        {
            sprintf(temp,"%d",sum);
            sum = 0;
            len = strlen(temp);
            for(i = 0;temp[i];i++)sum+=temp[i]-48;
        }
        printf("%d\n",sum);
    }
    return 0;
}

这个程序应该不行，，在输入38的时候，即sum大于9的时候，还是会输出sum，再去计算，，，这样就不符合题意了

这个也过不了！

我的能过

用一个函数，比如int fun （int n），当输入n小于等于9时，返回n，输入n大于9时，把n的每一位都加起来再调用fun函数，知道所有位数的和为个位数为止，具体实现如下：

int fun(int n)                        //函数保证返回值是基本数字
{
    int r = 0;
    if (n <= 9)
        return n;

    while (n > 0)
    {
       r += n%10;
       n /= 10;
    }
    return fun(r);
}

主函数中处理输入，把输入转换成一个整数n再进行函数fun的调用，基本上思想都这样。