矩阵加速（数列）

P1962 斐波那契数列

\(F_n = \begin{cases} 1 & (n \leq 2)\\ F_{n-1} + F_{n-2} & (n \geq 3 )\end{cases}\)

请你求出 \(F_n \text{ mod } 10^9+7\) 的值。

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\(\begin{cases} F_{n+1} = 1 × F_{n} + 1 × F_{n-1} \\ F_n = 1× F_n + 0× F_{n - 1} \end{cases}\)

\( \Rightarrow \left[\begin{array}{ccc} F_{n+1}\\ F_n \end{array}\right] = \left[\begin{array}{ccc} 1 & 1\\ 1 & 0 \end{array}\right] × \left[\begin{array}{ccc} F_n\\ F_{n- 1} \end{array}\right] \)

\( \Rightarrow \left[\begin{array}{ccc} F_{n+1}\\ F_n \end{array}\right] = \left[\begin{array}{ccc} 1 & 1\\ 1 & 0 \end{array}\right] ^ {n - 1} × \left[\begin{array}{ccc} F_2\\ F_1 \end{array}\right] \)

\( \Rightarrow \left[\begin{array}{ccc} F_{n+1}\\ \color{red}{F_n} \end{array}\right] = \left[\begin{array}{ccc} 1 & 1\\ 1 & 0 \end{array}\right] ^ {n - 1} × \left[\begin{array}{ccc} 1\\ 1 \end{array}\right] \)

则初始化矩阵为 \(\left[\begin{array}{ccc}1\\1\end{array}\right]\)，快速幂 \(n-1\) 次转移矩阵 \(\left[\begin{array}{ccc} 1 & 1\\ 1 & 0 \end{array}\right]\)，得到的 \(Ans_{2, 1}\) 即为答案。

创建

struct STU{
    ll m[3][3];
    STU(){memset(m, 0, sizeof m);}
    STU operator*(const STU &b) const{
        STU x;
        for(int i = 1; i <= 2; ++i)
            for(int j = 1; j <= 2; ++j)
                for(int k = 1; k <= 2; ++k)
                    x.m[i][j] = (x.m[i][j] + m[i][k] * b.m[k][j]) % mod;
        return x;
    }
} Ans, Base;

初始化

void Pre()
{
    Ans.m[1][1] = Ans.m[2][1] = 1;
    Base.m[1][1] = Base.m[1][2] = Base.m[2][1] = 1;
}

计算

void ksm(ll b)
{
    while(b)
    {
        if(b & 1)
            Ans = Ans * Base;
        Base = Base * Base;
        b >>= 1;
    }
}

int main()
{
    ll n = read();
    Pre();
    ksm(n - 1);
    printf("%lld\n", Ans.m[2][1] % mod);
    return 0;
}

标签：begin,right,end,数列,矩阵,array,加速,ccc,left
来源： https://www.cnblogs.com/EdisonBa/p/15292258.html