hdu1013 digital root（数根）

做了之后才知道可以使用数论做这个题
树根公式: d=(n-1)%9+1;

注意数字可能非常大，基本类型无法存储，使用字符数组存

Digital Roots
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 59431 Accepted Submission(s): 18589

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
数论过————————————————————————–

#include <cstdio>
#include <cstring>


int main(){
   char str[10000];
    while(scanf("%s",str)!=EOF){
        if((str[0]-'0')==0) break;
        int n=0;
        for(int i=0;i<strlen(str);i++)
            n+=str[i]-'0';
        printf("%d\n",(n-1)%9+1);
    }
    return 0;
}

模拟过——————————————————————-

#include <cstdio>
#include <cstring>


int main(){
   char str[10000];
    while(scanf("%s",str)!=EOF){
        if((str[0]-'0')==0) break;
        int n=0;
        for(int i=0;i<strlen(str);i++)
            n+=str[i]-'0';
       int ans=n;
       while(ans>=10){
            n=ans; ans=0;
            while(n){
                ans+=n%10;
                n/=10;
            }
       }
        printf("%d\n",ans);
    }
    return 0;
}