题目描述
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.
输入描述:
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.

The integer may consist of a large number of digits.
输出描述:
For each integer in the input, output its digital root on a separate line of the output.
示例1
输入
24
39
0
输出
6
3
解题思路：//2的31次方约等于2乘以10的9次方，即十进制最多十位数
//数位分解以后，就算是10个9的数位和也才90
//所以第一步处理以后，就只剩两位数了，循环计算两位数就行 。

#include<cstdio>
#include<cstring>
int main(){
    char str[35];
    while(scanf("%s",str)!=EOF&&str[0] != '0'){
        int len = strlen(str);
        int sum = 0;
        for(int i = 0;i < len;i++){
            sum += str[i] - '0';
        } 
        while(1){
            if(sum < 10){
                printf("%d\n",sum);
                break;
            }
            else {
                sum = sum/10 + sum%10; 
            }
        }

    }
    return 0; 
}