矩阵快速幂与快速斐波那契数列

矩阵快速幂与快速斐波那契数列

已知$f(n)=af(n-1)+bf(b-2)$，因为有两项所以我们构造一个$2\ast 2$的矩阵使得

$$\left[\begin{array}{cc}f(n-1)& f(n-2)\end{array}\right]\ast \left[\begin{array}{cc}t1& t2\\ t3& t4\end{array}\right]=\left[\begin{array}{cc}f(n)& f(n-1)\end{array}\right]$$

根据矩阵乘法规律

$$\begin{gathered} f(n)=t1\ast f(n-1)+t3\ast f(n-2) \\ f(n-1)=t2\ast f(n-1)+t4\ast f(n-2) \end{gathered}$$

推得$t1=a,t2=1,t3=b,t4=0$

即

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

所以有

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

同理斐波那契数列为

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

矩阵乘法：

#define ll long long
const int N = 3,mod = 1e9 + 7;
struct jz{
    ll a[N][N];
    jz(){memset(a,0,sizeof(a));}//构造
    jz(bool f){memset(a,0,sizeof(a));//单位矩阵
    for(int i = 1;i <= 2;i++)
        a[i][i] = 1;
    }//构造单位矩阵
    jz operator*(const jz &other)const{
        jz ans;
        for(int i = 1;i <= 2;i++)
            for(int j = 1;j <= 2;j++)
                for(int k = 1;k <= 2;k++)
                    ans.a[i][j] = (ans.a[i][j] + a[i][k]*other.a[k][j]) % mod;
       	return ans;
    }
}

单位矩阵：

从左上角到右下角的对角线上的元素均为1。除此以外全都为0。根据单位矩阵的特点，任何矩阵与单位矩阵相乘都等于本身

矩阵快速幂：

jz jzfpow(jz a,ll n){
	jz ans(true);
    while(n){
        if(n&1) ans = (ans*a);
        n >>= 1;
        a=a*a;
    }
    return ans;
}

const ll N = 101,mod = 1e9 + 7;
int n;
struct jz{
    ll a[N][N];
    jz(){memset(a,0,sizeof(a));}//构造
    jz(bool f){memset(a,0,sizeof(a));//单位矩阵
    for(int i = 1;i <= n;i++)
        a[i][i] = 1;
    }//构造单位矩阵
    jz operator*(const jz &other)const{
        jz ans;
        for(int i = 1;i <= n;i++)
            for(int j = 1;j <= n;j++)
                for(int k = 1;k <= n;k++)
                    ans.a[i][j] = (ans.a[i][j] + (a[i][k]*other.a[k][j]) % mod) % mod;
       	return ans;
    }
};
jz jzfpow(jz a,ll n){
	jz ans(true);
    while(n){
        if(n&1) ans = (ans*a);
        n >>= 1;
        a=a*a;
    }
    return ans;
}