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;}