矩阵快速幂求菲波那切数列

#include<cstdio>#include<cstring>using namespace std;struct Matrix{long long x[2][2];friend Matrix operator*(const Matrix &a,const Matrix &b){Matrix temp;temp.clear();for(int i=0;i<=1;i++)    for(int j=0;j<=1;j++)        for(int k=0;k<=1;k++){            temp.x[i][j]+=a.x[i][k]*b.x[k][j];            temp.x[i][j];}return temp;}}ans={2,1,1,1},inc={1,1,1,0};int main(){long long m;scanf("%lld",&m);m-=3;while(m){    if(m&1)        ans=ans*inc;    inc=inc*inc;    m>>=1;}printf("%lld",ans.x[0][0]);return 0;}