剑指 Offer 10- I. 斐波那契数列 —— C++矩阵快速幂求解

解题思路

转移方程：f(n)=f(n-1)+f(n-2),f(n-1)=f(n-1)

$$\left[\begin{array}{c}f(n)\\ f(n-1)\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]\left[\begin{array}{c}f(n-1)\\ f(n-2)\end{array}\right]$$

代码

class Solution {
public:
    long* mul(const long* a,const long* b,const int row,const int col,const int kmax) const{
        
        long* ans=new long[row*col];

        for(int i = 0;i<row;i++){
            for(int j = 0;j<col;j++){
                ans[i*col+j]=0;
                for(int k = 0;k<kmax;k++){
                    ans[i*col+j]+=(a[i*kmax+k]*b[k*col+j])%(1000000007);
                }
            }
        }
        return ans;
    }

    int fib(int n) {
        if(n==0) return 0;
        if(n==1||n==2){
            return 1;
        }
        long* coe=new long[]{1,1,1,0};
        long* ans=new long[]{1,1};
        n-=2;
        while(n){
            if(n&1)
                ans = mul(coe,ans,2,1,2);
            coe = mul(coe,coe,2,2,2);
            n>>=1;
        }
        return ans[0]%1000000007;
    }
};