杭电ACM 1013 Digital Roots

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
问题分析与算法设计
题目要求求出一个整数的根，并且整数可能无限大，所以考虑用字符串来保存数字，这样可以使各位数字相对独立，且每个数字范围为0~9，首先将字符串数组每一项转换成数字形式，这里用到了sum+=str[i]-‘0’，然后求出第一次各位数字相加的和。用sum模除10读出个位数位，然后用整除10得到sum减去个位的值，再将两者作和，如此循环，终止条件是sum模除10读出个位数位等于自身，这就表明此时sum的值即为该整数的根。
源码如下：
[code]#include <iostream>
#include <string>
using namespace std;

int main()
{
	string str;
	while(cin>>str&&str!="0") 
	{
		int sum=0;
		for(int i=0;i<str.length();i++)
			sum+=str[i]-'0';
			while(sum%10!=sum)
			{
				int a=sum%10;
				int b=sum/10;
				sum=a+b;
			}
			cout<<sum<<endl;
	}
	return 0;
}

测试结果截图

[/code]