Java - Nuc - 郭姐散步 ( 矩阵快速幂 )

描述

The first lady of our team is GuoJie, GuoJie likes walking very much.

Today GuoJie walks from the original point (0, 0), everytime he(may be she?) can go up or left or right a step.

But she can't go back the point where she have visited.

For example, if he goes up a step, she will be at (1, 0) and she never comes back the point.

Now, if she can walk n(n <= 1000000) steps, can you find how many ways she can walk? the result mod 1e9 + 7. 

输入 There will be T (T <= 100) cases, each case will input a n. 输出 For each group of input integers you should output how many ways GuoJie can walk in one line, and with one line of output for each line in input. 样例输入

1
2

样例输出

7

提示 n<=0 输出0 来源

Json

由分析可知，每一步的下一步可以有两种或者三种选择，有两种选择的步其下一步的选择数为2+3，有三种选择的步其下一步为2+2+3.

所以设a_n为2的个数，b_n 为3的个数，ans即为2*a_n+3*b_n 

import java.util.Scanner;
public class Main{
	final static int Mod = (int)1e9+7;
	public static void main(String[]args){
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		for(int i=1 ; i<=T ;i++){
			int n = sc.nextInt();
			
			if(n<0||n==0){
				System.out.println(0);
			}else if(n==1){
				System.out.println(3);
			}else{
				long[][]a = new long[2][1];
				a[0][0] = 0;
				a[1][0] = 1;
				long[][]c = new long[2][2];
				c[0][0] = 1;
				c[0][1] = 2;
				c[1][0] = 1;
				c[1][1] = 1;
				
				long[][]x = new long[2][1];
				x = Multiply_Matrix(Matrix_Ksm(c,n-1),a);
				long times = (x[0][0]*2%Mod+x[1][0]*3%Mod)%Mod;
				System.out.println(times);	
			}
			
		}
	}
	public static long[][] Multiply_Matrix(long[][]a ,long[][]b){		//矩阵乘法
		long[][]c = new long [a.length][b[0].length];
		for(int i=0 ;i

有算法优化请评论