Hduoj1163 【数学】【水题】

/*Problem D 
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 25   Accepted Submission(s) : 11
Font: Times New Roman | Verdana | Georgia 
Font Size: ← →
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>
int main()
{
	__int64 i, j, k, a, b, c;
	while(scanf("%I64d", &a)  != EOF && a)
	{
		b = a;
		c = a;
		while(--a)
		{
			b %= 9;
			b *= c;
		}
		b %= 9;
		if(b)
		printf("%I64d\n", b);
		else
		printf("9\n"); 
	}
	return 0;
}

题意：求一个数n的n次方的各位数之和的根。

思路：将 n对 9取余， 得到的结果即是n各位数之和， 再将得到的结果乘以n，重复执行n次，所得结果即是答案。

关键：n的n次方   的解  可以  拆分成 单个n的解乘以n的n-1次方。。。。