HDU1013_Digital Roots

Digital Roots

*Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 76568 Accepted Submission(s): 23909*

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

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今天做了几道很简单的水题，AC的感觉很棒！

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这道题个人觉得需要注意一点：就是用int或者long long 都会WA，因为输入的数字可能会很长，改用成字符数组存储。

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#include 
#include 
#include 
using namespace std;

char array[1000005];

int main()
{
    //long long n;
    while((scanf("%s",array)))
    {
        if(array[0]-'0'==0)
            break;
        int sum = 0;
        for(int i = 0; i < strlen(array); i++)
        {
            sum += array[i]-'0';
            if(sum>9)
                sum = sum%10 + sum/10; //sum最大值也就是18
        }
        printf("%d\n",sum);
    }
    return 0;
}