矩阵求斐波那契数列第n项

记一个板子 原博客在这

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#include<bits/stdc++.h>

using namespace std;
const int MAX = 10;
#define __int64 long long
#define Bit(n) 1<<n
#define CLR(arr, val) memset(arr,val,sizeof(arr))
const int mod = 1e9 + 7;

class Matrix {
public:
    Matrix(int r, int c) : row(r), col(c) {}

    void Init() {
        CLR(map, 0);
        map[0][0] = map[0][1] = map[1][0] = 1;
    }

    void Unit() //初始化为单位矩阵
    {
        CLR(map, 0);
        for (int i = 0; i < row; i++)
            map[i][i] = 1;
    }

    int Result() const { return map[0][1] % mod; }

    friend Matrix operator*(const Matrix &, const Matrix &);

    int Pow(int);

private:
    __int64 map[MAX][MAX];
    int row, col;
};

Matrix operator*(const Matrix &M1, const Matrix &M2) //矩阵相乘模板
{
    Matrix M(M1.row, M2.col); //相乘之后矩阵的行和列会变化
    for (int i = 0; i < M1.row; i++)
        for (int j = 0; j < M2.col; j++) {
            M.map[i][j] = 0;
            for (int k = 0; k < M1.col; k++)
                M.map[i][j] += M1.map[i][k] * M2.map[k][j];
            M.map[i][j] %= mod;
        }
    return M;
}

Matrix M(2, 2);

int Matrix::Pow(int n) //矩阵快速幂
{
    Matrix temp(2, 2);
    temp.Init();
    for (int i = 0; Bit(i) <= n; i++) //利用二进制的思想求解
    {
        if (Bit(i) & n) M = M * temp;
        temp = temp * temp;
    }
    return M.Result();
}

int main() {
    int k;
    scanf("%d", &k);
    while(k--){
        __int64 num;
        cin >> num;
        M.Unit();
        cout << M.Pow(num) << endl;
    }
    return 0;
}

​

再搞一个矩阵的板子

struct Matrix {
    ll a[maxn][maxn];

    Matrix operator*(const Matrix &b) const {
        Matrix c;
        for (int i = 1; i <= k; i++) {
            for (int j = 1; j <= k; j++) {
                c.a[i][j] = 0;
                for (int u = 1; u <= k; u++) {
                    c.a[i][j] = (c.a[i][j] + a[i][u] * b.a[u][j] % (mod - 1)) % (mod - 1);
                }
            }
        }
        return c;
    }

    Matrix pow(ll x) const {
        Matrix b = *this, r = *this;
        x--;
        while (x > 0) {
            if (x & 1) r = r * b;
            b = b * b;
            x >>= 1;
        }
        return r;
    }
};