HDu 1005 Number Sequence(数論)

Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 86102 Accepted Submission(s): 20423
Problem Description
A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
 
 
Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
 
 
Output
For each test case, print the value of f(n) on a single line.
 
 
Sample Input
1 1 3 1 2 10 0 0 0
 
 
Sample Output
2 5
 
 
Author
CHEN, Shunbao
 
 
Source
ZJCPC2004  
その时、他の人は私にこの问题を闻いて、私は最初は行列の速いべき乗だと感じて、帰ってから%7を考えて、7が小さすぎて、规则があるかもしれないので、周期は最大48で、突然元の前にこの问题をしたことがあることを思い出します

#include <stdio.h>
int main()
{
	int i,j,A,B,f1,f2,f3,n;
	while(scanf("%d%d%d",&A,&B,&n)!=EOF&&(A||B||n))
	{
		f1=1;
		f2=1;
		A=A%7;
		B=B%7;
		n=(n-3)%48+2;
		for(i=2;i<=n;i++)
		{
			f3=(A*f2+B*f1)%7;
			f1=f2;
			f2=f3;
		}
		printf("%d
",f2);
	}
	return 0;
}