HDU1013 Digital Roots

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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：给一串数字，使每个位上的数字相加，如果结果大于等于10 ，循环这个过程，知道结果小于十，输出。

      实际上是个大数问题，给出的数字可能已超过了LONG LONG 范围，可以用字符串处理。

 

#include<cstring>
#include<cstdio>
#include<iostream>
#include<cstdlib>

using namespace std;

int main()
{
    char str[20000];
    while(cin>>str&&str[0]!='0')
    {
        int sum;
        do{
                sum=0;
            for(int i=0;i<strlen(str);i++)
                sum+=(str[i]-'0');
            itoa(sum,str,10);  //将数字转换为字符串 itoa（数字，字符串首地址，进制）；
        }while(sum>=10);
        cout<<sum<<endl;
    }
}