矩阵快速幂模板(重载运算符)

题意

求斐波那契数列第n项，n<=1e18

solution

1.显然O(n)递推肯定不行
2.所以我们考虑用矩阵快速幂加速递推

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#include<ctime>
using namespace std;
typedef long long ll;
const ll inf=0x3f3f3f3f;
const ll p=100000007;
inline ll read(){
	char ch=' ';ll f=1;ll x=0;
	while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
struct Matrix
{
	ll x,y;
	ll m[3][3];
}a,b,c;
Matrix operator *(Matrix a,Matrix b)
{
	ll n,m,q;
	n=a.x;m=a.y;
	q=b.y;
	Matrix tmp;
	tmp.x=n;tmp.y=q;
	memset(tmp.m,0,sizeof(tmp.m));
	for(ll i=1;i<=n;i++)
	{
		for(ll j=1;j<=m;j++)
		{
			for(ll k=1;k<=q;k++)
			{
				tmp.m[i][j]+=a.m[i][k]*b.m[k][j];
				tmp.m[i][j]%=p;
			}
		}
	}
	return tmp;
}
Matrix mul(Matrix x,ll y)
{
	Matrix ret;
	ret.m[1][1]=1;
	ret.m[1][2]=0;
	ret.m[2][1]=0;
	ret.m[2][2]=1;
	ret.x=2;ret.y=2;
	while(y)
	{
		if(y&1)
		{
			ret=ret*x;
		}
		x=x*x;
		y=y>>1;
	}
	return ret;
}
int main()
{
	ll n=read();
	a.m[1][1]=1;
	a.m[1][2]=1;
	a.m[2][1]=1;
	a.m[2][2]=0;
	a.x=2;a.y=2;
	
	b.m[1][1]=1;
	b.m[1][2]=1;
	b.x=1;b.y=2;
	c=mul(a,n);
	b=b*c;
	cout<<b.m[1][1]<<endl;
	return 0;
}