hdu 1163 Eddy's digital Roots(九余數定理)

hdu 1163 Eddy's digital Roots

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
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看FFT看的頭暈還是沒看懂，然後回頭刷水題，想找點快感，沒想到碰到個小炸彈。。。

用到了一個九余數定理，不知道這個定理確實很難做……

九余數定理簡介：一個數N 各位數字的和 對9 取余等於 這個數對 9取余。。

太坑了，記下來，免得以後忘了……

**********************/

#include 
#include 
using namespace std;
int main()
{
    int n,i,s;
    while(scanf("%d",&n)&&n)
    {
        s = 1;
        for(i = 1;i<=n;i++){
            s=s*n%9;
        }
        if(s==0)    printf("9\n");
        else    printf("%d\n",s);
    }
    return 0;
}

太坑了……