Leetcode 509. 斐波那契数

利用矩阵来解，当然可以用递归，递推去做

如图：

所以f(n)就是左边矩阵的n-1次方后第1行第一列的数字

class Solution {
public:
	struct Matrix//矩阵储存
	{
		long long mat[3][3];

	};
	Matrix e;//单位矩阵
	
	Matrix Mul(Matrix a, Matrix b)//矩阵相乘
	{
		Matrix temp;
		for(int i = 1; i <= 2; i++)
			for(int j = 1; j <= 2; j++)
				temp.mat[i][j] = 0;
		for(int i = 1; i <= 2; i++)
		{
			for(int j = 1; j <= 2; j++)
			{
				for(int k = 1; k <= 2; k++)
				{
				temp.mat[i][j] += a.mat[i][k] * b.mat[k][j];
				}
			}
		}
		return temp;
	}

	

	Matrix quickpow(Matrix a, int N)//快速幂
	{
		for(int i = 1; i <= 2; i++)//初始单位矩阵
			e.mat[i][i] = 1;
		Matrix res = e;
		while(N)
		{
			if(N & 1)
				res = Mul(res, a);
			a = Mul(a, a);
			N>>=1;
		}
		return res;

	}
    int fib(int N) {
        if(N<=0)
			return 0;
		if(N == 1)
			return 1;
		Matrix st;
		st.mat[1][1] = 1;
		st.mat[1][2] = 1;
		st.mat[2][1] = 1;
		st.mat[2][2] = 0;
		Matrix result = quickpow(st, N);
		return result.mat[1][1];
    }
};

 

来源：https://blog.csdn.net/qq_40722582/article/details/97310256