HDU 1163 Eddy's digital Roots

计算方法是：这个数字根等于原数除以9的余数，因此这个计算过程常常称为“合九法”。如39/9=3.成立

同余定理：如果两个乘积除以m的余数等于这两个数分别除以m的余数积。 例如：7%3=1 5%3=2 7*5/3=2=1*2

Eddy's digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2759    Accepted Submission(s): 1592

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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JGShining

 

#include <iostream>
#include <stdlib.h>
using namespace std;
int n;
int main( )
{
    while( cin >> n  && n ){
           int sum = 1;
           for( int i=0 ; i<n ; i++ ){
                sum = sum * n % 9;     
           }
           if( sum == 0 )
               cout << 9 << endl;
           else
               cout << sum << endl;
    }
    return 0;
}