HDOJ 1013

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30962    Accepted Submission(s): 9508


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

刚开始忽略的数的大小，代码AC

#include <iostream>
using namespace std;
void sum(int n);
int main()
{
    int n;
    while(cin>>n)
    {
        if(n == 0)
            break;
        else
        {
            sum(n);
        }
    }
    return 0;
}
void sum(int n)
{
    int count=0;
    while(n>=10)
    {
        count+=n%10;
        n=n/10;
    }
    count+=n;
    if(count>10)
        sum(count);
    else
        cout<<count<<endl;
}

后来改用按位来做....需要判断是否为0时候，要注意10这种数，最后一位为0也需要判断...因此改用sum是否为0来确定，是否是第一个数字为0

#include <iostream>
using namespace std;
int main()
{
    char c=' ';
    int n,sum=0;
    while(c=getchar())
    {
        if(sum>=10)
                sum=sum%10+sum/10;
        if(c == '\n')
        {
            if(sum==0)
                break;
            cout<<sum<<endl;
            sum=0;
        }
        else
        {
            n=c-48;
            sum+=n;    
        }
    }
    return 0;
}