POJ 1519：Digital Roots

Digital Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25766		Accepted: 8621

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

题意是给出一个数，求每一位的和，如24，就是2+4=6。如果得到的数还是大于等于10的数，就对这个数继续求每一位的和，直到变成个位数。如39>12->3。

不能更标准的递归了。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
using namespace std;

string out;

void recursion(string test)
{
	if(test.length()==1)
	{
		out=test;
		return;
	}
	int i,result=0;
	for(i=0;i<test.length();i++)
	{
		result += test[i]-'0';
	}
	int temp;
	string test2;
	while(result!=0)
	{
		temp=result%10;
		test2 += temp+'0';
		result /=10;
	}
	recursion(test2);
}

int main()
{
	string test;

	while(cin>>test)
	{
		if(test=="0")
			break;
		recursion(test);
		cout<<out<<endl;
	}
	return 0;
}

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