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hdu1013 Digital Roots (模拟||九餘數定理)

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

可以九餘數定理,以為可以像我們下面那樣模拟一下。

#include<cstdio>
#include<cstring>
/*
九餘數定理
一個數對九取餘後的結果稱為九餘數。
一個數的各位數字之和想加後得到的<10的數字稱為這個數的九餘數(如果相加結果大于9,則繼續各位相加)
*/
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 ;
}