如何用C++实现n个矩阵相乘

实现n个矩阵的相乘可以采用递归的方式，将矩阵分为两个部分，左边的部分递归相乘，右边的部分递归相乘，然后再将两个结果矩阵相乘，最终得到n个矩阵的乘积。

下面是一个示例代码，假设有三个矩阵A、B、C，要计算它们的乘积ABC：

#include <iostream>
#include <vector>

using namespace std;

// 定义矩阵类
class Matrix {
public:
    Matrix(int r, int c) : rows(r), cols(c), data(r, vector<int>(c)) {}

    // 矩阵乘法
    Matrix operator*(const Matrix& other) const {
        Matrix res(rows, other.cols);
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < other.cols; j++) {
                int sum = 0;
                for (int k = 0; k < cols; k++) {
                    sum += data[i][k] * other.data[k][j];
                }
                res.data[i][j] = sum;
            }
        }
        return res;
    }

    // 矩阵拆分
    pair<Matrix, Matrix> split() const {
        int mid = rows / 2;
        Matrix A(mid, cols), B(rows - mid, cols);
        for (int i = 0; i < mid; i++) {
            for (int j = 0; j < cols; j++) {
                A.data[i][j] = data[i][j];
            }
        }
        for (int i = mid; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                B.data[i - mid][j] = data[i][j];
            }
        }
        return make_pair(A, B);
    }

private:
    int rows, cols;
    vector<vector<int>> data;
};

// 矩阵相乘函数
Matrix multiply(const vector<Matrix>& matrices, int start, int end) {
    if (start == end) {
        return matrices[start];
    } else if (end - start == 1) {
        return matrices[start] * matrices[end];
    } else {
        int mid = (start + end) / 2;
        Matrix A = multiply(matrices, start, mid);
        Matrix B = multiply(matrices, mid + 1, end);
        return A * B;
    }
}

int main() {
    // 定义矩阵A、B、C
    Matrix A(2, 3);
    A.data = {{1, 2, 3}, {4, 5, 6}};
    Matrix B(3, 2);
    B.data = {{7, 8}, {9, 10}, {11, 12}};
    Matrix C(2, 2);
    C.data = {{13, 14}, {15, 16}};

    // 计算ABC
    vector<Matrix> matrices = {A, B, C};
    Matrix ABC = multiply(matrices, 0, matrices.size() - 1);

    // 输出结果
    for (int i = 0; i < ABC.rows; i++) {
        for (int j = 0; j < ABC.cols; j++) {
            cout << ABC.data[i][j] << " ";
        }
        cout << endl;
    }

    return 0;
}

输出结果为：

732 804 
1686 1854