HDU 1163 Eddy's digital Roots（九余数定理）

Eddy’s digital Roots

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5860 Accepted Submission(s): 3232

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.

The Eddy’s easy problem is that : give you the n,want you to find the n^n’s digital Roots.

Input
The input file will contain a list of positive integers n, one per line. The end of the input will be indicated by an integer value of zero. Notice:For each integer in the input n(n<10000).

Output
Output n^n’s digital root on a separate line of the output.

Sample Input
2
4
0

Sample Output
4
4

Author
eddy

题解：（九余数定理）

一个数对九取余，得到的数称之为九余数；

一个数的九余数等于它的各个数位上的数之和的九余数。

AC代码：

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>#include<algorithm>using namespace std;int main(){    int n,m;    while(~scanf("%d",&n)){        if(n==0)break;        m=n;        for(int i=1;i<n;i++){            m=(m*n)%9;        }        if(m==0)printf("%d\n",9);        else printf("%d\n",m);    }   return 0;}