ZOJ_1115_Digital Roots

Digital Roots

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Background

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.

Example

Input

24
39
0

Output

6
3

     

题意和思路：题目就是定义了一个概念树根，它必须是一位的，得到方式是各个数位的和，因为要求是一位所以如果不是一位，则重复这样的流程，即继续将各个位数相加，直到得到树根为止。这题理解题意，做起来很简单。

注意点：一般来说第一遍很大可能是错误的，就是超时了，因为这里可能是很大的整数，所以第一次处理的时候将其看成字符串，然后字符串的各个位数相加，之后再按整数的方式处理就不会超时的，我一开始也错了。。。简单题也不要掉以轻心。

Sample Program Here

#include<iostream>
#include<string>
using namespace std;


//////用来得到各个位数的和
int process(int a){
	int sum=0;
	while(a!=0){
	  sum+=a%10;
	  a/=10;
	}

	return sum;
}

int main(){
 int root=0;  //a是输入的数，root 是相应的数根
 string a;
 while(cin>>a){
    root=0;
    if(a.length()==1&&a[0]=='0') break;

	//得到第一次计算的各位数和
	for(int i=0;i<a.length();i++){
		root+=a[i]-'0';
	}


    //如果处理后不是一位的则继续操作
	while(root/10!=0){   
	   root=process(root);
	}

	cout<<root<<endl;

 }
 return 0;
}