洛谷P1962 斐波那契数列 题解 矩阵快速幂

题目链接：https://www.luogu.com.cn/problem/P1962

解题思路：

因为

\[\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \times \begin{bmatrix} F_{i} \\ F_{i+1} \end{bmatrix} = \begin{bmatrix} F_{i+1} \\ F_{i+2} \end{bmatrix} \]

所以

\[\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}^{n-2} \times \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} = \begin{bmatrix} F_{n-1} \\ F_{n} \end{bmatrix} \]

示例程序：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 5;
const long long mod = 1e9 + 7;

struct Matrix {
    int n, m;
    long long a[maxn][maxn];

    void init(int _n, int _m) {
        n = _n;
        m = _m;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                a[i][j] = 0;
    }

    Matrix operator * (const Matrix &b) const {
        Matrix res;
        res.init(n, b.m);
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                for (int k = 1; k <= b.m; k++)
                    res.a[i][k] += a[i][j] * b.a[j][k],
                    res.a[i][k] %= mod;
        return res;
    }

    Matrix operator ^ (long long k) const {
        Matrix res, t;
        res.init(n, n);
        for (int i = 1; i <= n; i++)
            res.a[i][i] = 1;
        t.init(n, n);
        memcpy(t.a, a, sizeof a);
        for (; k; k >>= 1, t = t * t)
            if (k & 1ll)
                res = res * t;
        return res;
    }
} a, b;
long long n;

int main() {
    cin >> n;
    if (n <= 2) {
        cout << 1 << endl;
        return 0;
    }
    a.init(2, 2);
    a.a[1][2] = 1;
    a.a[2][1] = a.a[2][2] = 1;
    a = a ^ (n-2);
    b.init(2, 1);
    b.a[1][1] = b.a[2][1] = 1;
    a = a * b;
    cout << a.a[2][1] << endl;
    return 0;
}