数列第N项 20 (矩阵 + 快速幂)

题目： 传送门
题意： 已知数列递推公式为
$$a_n=(3a_{n-1}+4a_{n-2})mod1e9+7$$
现在给你a1和a2，要求$a_n$
思路：
看如下矩阵乘法：
$$\left(\begin{array}{c}{a}_n\\ a_{n-1}\end{array}\right)=\left(\begin{array}{cc}3& 4\\ 1& 0\end{array}\right)\ast \left(\begin{array}{c}{a}_{n-1}\\ a_{n-2}\end{array}\right)$$
可以一直递推，得到：
$$\left(\begin{array}{c}{a}_n\\ a_{n-1}\end{array}\right)={\left(\begin{array}{cc}3& 4\\ 1& 0\end{array}\right)}^{n-1}\ast \left(\begin{array}{c}{a}_2\\ a_1\end{array}\right)$$
如果
$${\left(\begin{array}{cc}3& 4\\ 1& 0\end{array}\right)}^{n-1}=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$$
那$a_n$ = $A\ast a_2+B\ast a_1$
由于n到了1e12，所以用快速幂
Code：

#pragma GCC optimize(2)
#include<iostream>
using namespace std;

typedef long long ll;
const ll mod = 1000000007;
struct matrix {
    
    
	ll a, b, c, d;
	matrix() {
    
    }
	matrix (ll aa, ll bb, ll cc, ll dd) {
    
    
		a = aa;
		b = bb;
		c = cc;
		d = dd;
	}
	matrix operator * (const matrix& x) {
    
    
		matrix tmp;
		tmp.a = a*x.a + b*x.c;
		tmp.b = a*x.b + b*x.d;
		tmp.c = c*x.a + d*x.c;
		tmp.d = c*x.b + d*x.d;
		return tmp;
	}
	matrix operator % (ll k) {
    
    
		matrix tmp(a%k, b%k, c%k, d%k);
		return tmp;
	}
};
matrix qpow(matrix a, ll n) {
    
    
	matrix ans(1,0,0,1);
	matrix t = a;
	while (n > 0) {
    
    
		if (n & 1) {
    
    
			ans = ((ans)%mod * (t%mod)) % mod;
		}
		t = ((t%mod) * (t%mod)) % mod;
		n >>= 1;
	}
	return ans % mod;
}
int main()
{
    
    
	ll n, a1, a2;
	cin>>n>>a1>>a2;
	matrix a(3,4,1,0), c;
	c = qpow(a, n - 2);
	//cout<<c.a<<" "<<c.b<<endl<<c.c<<" "<<c.d<<endl;
	cout<<(c.a * (a2%mod) + c.b * (a1 % mod)) % mod<<endl;
	return 0;
}