hdu 1163 Eddy's digital Roots （数根）

Eddy's digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3351    Accepted Submission(s): 1897

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

240

 

Sample Output

44

 

Author

eddy

 

Recommend

JGShining

 

题解：这是一道经典的数根问题，首先要知道一个数论知识点，一个非零n进制数mod（n-1）的结果，若为0修正为n-1，这个结果和原来数的数根一样，证明略。

知道上述结论这道题目就很好做了，AC代码如下（c语言）：

//不用快速幂的代码#include <stdio.h>#include <stdlib.h>int main(){    int n,i,res,mul;    while(scanf("%d",&n),n)    {        for(i=0,res=1,mul=n%9;i<n;i++)            res=mul*res%9;        res=res==0?9:res;        printf("%d\n",res);    }    return 0;}//用快速幂的代码，由于n很小，不用都能过#include <stdio.h>#include <stdlib.h>int POW(int a,int b){    int res=1,mul=a;    while(b)    {        if(b&1) res=mul*res%9;        mul=mul*mul%9;        b=b>>1;    }    return res;}int main(){    int n,res,m;    while(scanf("%d%d",&n,&m),n&&m)    {        res=POW(n,m);        res=res==0?9:res;        printf("%d\n",res);    }    return 0;}