Number Sequence_hdu_1005(法则)

Number Sequence_hdu_1005(规律)
9329854 2013-10-13 14:36:41 Accepted 1005 171MS 5072K 654 B Java zhangyi
http://acm.hdu.edu.cn/showproblem.php?pid=1005
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 86162    Accepted Submission(s): 20434


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

*/

import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		Scanner input=new Scanner(System.in);
		while(true){
			int a=input.nextInt();
			int b=input.nextInt();
			long n=input.nextInt();
			if(a==0&&b==0&&n==0)
				break;
			int f[]=new int[1000];
			f[1]=f[2]=1;
			int s[][]=new int[7][7];
			s[1][1]=2;
			int i=0;
			for(i=3;i<60;i++){
				f[i]=(a*f[i-1]+b*f[i-2])%7;
				if(s[f[i-1]][f[i]]!=0){
					break;
				}
				s[f[i-1]][f[i]]=i;
			}
			int d=i-s[f[i-1]][f[i]];
			n-=s[f[i-1]][f[i]];
			n%=d;
			if(n==0)n+=d;
			n+=s[f[i-1]][f[i]];
			System.out.println(f[(int)n]);
		}
	}
}