题目1124：Digital Roots (方法超简单)

题目1124：Digital Roots

学到的新知识

求一个数各个的和可以把其%9就行，例如13%9=4 11%9=2；123%9=6；

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：3819

解决：1335

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：

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.

输出：

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

样例输入：

24
39
0

样例输出：

6
3

提示：

The integer may consist of a large number of digits.

//题目中说数很大，可能真的很大，需要用字符数组来存储
#include <iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
using namespace std;

int main()
{
    int i,m;
    char s[100];
    while(scanf("%s",s)!=EOF&&s[0]!=‘0‘)
    {
        for(m=i=0;s[i];i++)
            m+=s[i]-‘0‘;
        printf("%d\n",m%9==0?9:m%9);//这个人是天才哈，求一个数各个的和可以把其%9就行，例如13%9=4 11%9=2；123%9=6；
    }
    return 0;
}