HDoj-1163- 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

#include<stdio.h>
#include<string.h>
int main()
{
	int n;
	while(~scanf("%d",&n),n)
	{
		int s=1;
		for(int i=0;i<n;i++)
		{
			s=s*n%9;       //事实上不难发现对9取余更简便。不解释为什么，仅仅能说这是一种规律

		}
		if(s==0)
		   printf("9\n");
		else
		   printf("%d\n",s);<pre name="code" class="cpp">

}return 0;}

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int sum_dig(int n)
{
	int m,sum=0;
	while(n)
	{
		m=n%10;
		sum+=m;
		n/=10;
	}
	return sum;
}
int main()
{
	int n;
	while(~scanf("%d",&n),n)
	{
		int s=1;
		for(int i=0;i<n;i++)
		{
		      s=n*sum_dig(s);
		}
		while(s>9)
		{
			s=sum_dig(s);
		}
		printf("%d\n",s);
	}
	return 0;
}