hdu1013 Digital Roots (模拟||九余数定理)

Digital Roots

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 69933 Accepted Submission(s): 21890

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

Sample Output

6

3

可以九余数定理，以为可以像我们下面那样模拟一下。

#include<cstdio>
#include<cstring>
/*
九余数定理
一个数对九取余后的结果称为九余数。
一个数的各位数字之和想加后得到的<10的数字称为这个数的九余数（如果相加结果大于9，则继续各位相加）
*/
char a[];
int sum[];

int main() {
    while(~scanf("%s",a)){
        int len=strlen(a);
        if(len==&&a[]=='0'){
            break;
        }
        long long sum=;
        for(int i=;i<len;++i){
            sum+=(a[i]-'0');
        }
        printf("%lld\n",sum%==?:sum%);
//        while(sum>=10){
//            long long n=sum;
//            sum=0;
//            while(n){
//                sum+=(n%10);
//                n/=10;
//            }
//        }
//        printf("%lld\n",sum);
    }
    return ;
}