hdu 1005 快速幂

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 93488    Accepted Submission(s): 22297

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

 

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import java.util.Scanner;


public class Main {

	static void print(int[][] m){
		System.out.println();
		for(int i=0; i<2; i++){
			for(int j=0; j<2; j++)
				System.out.print(m[i][j]+" ");
			System.out.println();
		}
	}
	
	static void mul(int[][] a, int[][] b){
		int[][] c = new int[2][2];
		for(int i=0; i<2; i++)
			for(int j=0; j<2; j++){
				c[i][j]=0;
				for(int k=0; k<2; k++)
					c[i][j] = (c[i][j] + a[i][k]*b[k][j])%7;
			}
		for(int i=0; i<2; i++)
			for(int j=0; j<2; j++) b[i][j]=c[i][j];
	}
	
	public static int quickPower(int n, int a, int b){
		int[][] m = {{1,0},{1,0}};
		int[][] g = {{a,b},{1,0}};
		while(n>0){
			if((n & 1) == 1) mul(g,m);
			mul(g,g);
			n>>=1;
		}
//		print(m);
		return m[0][0];
	}
	
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s = new Scanner(System.in);
		while(true){
			int a = s.nextInt(), b=s.nextInt(),n=s.nextInt();
			if(a==0 && b==0 && n==0) break;
			int ans = n < 3 ? 1 : quickPower(n-2, a, b);
			System.out.println(ans);
		}
	}
}