设A 和 B 是两个n * n阶矩阵，求它们的乘积矩阵C。要求使用分治法，请讲述算法设计的思路

好的，我明白了，下面是算法设计的思路：

首先，将矩阵A和矩阵B进行分割，分成四个n/2 * n/2的子矩阵。然后，对于这四个子矩阵，分别求它们的乘积矩阵。

接下来，我们可以使用分治法将这些子矩阵的乘积合并成最终的乘积矩阵C。具体方法是：

将C分成四个n/2 * n/2的子矩阵，分别表示为C11，C12，C21和C22。然后，通过公式C11 = A11 * B11 + A12 * B21，C12 = A11 * B12 + A12 * B22，C21 = A21 * B11 + A22 * B21，C22 = A21 * B12 + A22 * B22，计算出这些子矩阵。

最后，将这些子矩阵组合成最终的乘积矩阵C即可。

这样，我们就通过分治法快速地计算出了两个n * n阶矩阵的乘积矩阵C。

相关问题

设A和B是两个n*n阶的矩阵，用分治法求他们的乘积矩阵C。使用Strassen算法。C++代码

#include <iostream>
#include <vector>

using namespace std;

vector<vector<int>> strassen(vector<vector<int>>& A, vector<vector<int>>& B) {
    int n = A.size();
    vector<vector<int>> C(n, vector<int>(n));

    if (n == 1) {
        C[0][0] = A[0][0] * B[0][0];
        return C;
    }

    vector<vector<int>> A11(n/2, vector<int>(n/2)), A12(n/2, vector<int>(n/2)), A21(n/2, vector<int>(n/2)), A22(n/2, vector<int>(n/2));
    vector<vector<int>> B11(n/2, vector<int>(n/2)), B12(n/2, vector<int>(n/2)), B21(n/2, vector<int>(n/2)), B22(n/2, vector<int>(n/2));
    vector<vector<int>> S1(n/2, vector<int>(n/2)), S2(n/2, vector<int>(n/2)), S3(n/2, vector<int>(n/2)), S4(n/2, vector<int>(n/2)), S5(n/2, vector<int>(n/2)), S6(n/2, vector<int>(n/2)), S7(n/2, vector<int>(n/2)), S8(n/2, vector<int>(n/2)), S9(n/2, vector<int>(n/2)), S10(n/2, vector<int>(n/2));
    vector<vector<int>> P1(n/2, vector<int>(n/2)), P2(n/2, vector<int>(n/2)), P3(n/2, vector<int>(n/2)), P4(n/2, vector<int>(n/2)), P5(n/2, vector<int>(n/2)), P6(n/2, vector<int>(n/2)), P7(n/2, vector<int>(n/2));

    // 分解A、B
    for (int i = 0; i < n/2; i++) {
        for (int j = 0; j < n/2; j++) {
            A11[i][j] = A[i][j];
            A12[i][j] = A[i][j+n/2];
            A21[i][j] = A[i+n/2][j];
            A22[i][j] = A[i+n/2][j+n/2];

            B11[i][j] = B[i][j];
            B12[i][j] = B[i][j+n/2];
            B21[i][j] = B[i+n/2][j];
            B22[i][j] = B[i+n/2][j+n/2];
        }
    }

    // 计算S、P
    S1 = B12 - B22;
    P1 = strassen(A11, S1);

    S2 = A11 + A12;
    P2 = strassen(S2, B22);

    S3 = A21 + A22;
    P3 = strassen(S3, B11);

    S4 = B21 - B11;
    P4 = strassen(A22, S4);

    S5 = A11 + A22;
    S6 = B11 + B22;
    P5 = strassen(S5, S6);

    S7 = A12 - A22;
    S8 = B21 + B22;
    P6 = strassen(S7, S8);

    S9 = A11 - A21;
    S10 = B11 + B12;
    P7 = strassen(S9, S10);

    // 计算C
    vector<vector<int>> C11(n/2, vector<int>(n/2)), C12(n/2, vector<int>(n/2)), C21(n/2, vector<int>(n/2)), C22(n/2, vector<int>(n/2));
    C11 = P5 + P4 - P2 + P6;
    C12 = P1 + P2;
    C21 = P3 + P4;
    C22 = P5 + P1 - P3 - P7;

    // 合并C
    for (int i = 0; i < n/2; i++) {
        for (int j = 0; j < n/2; j++) {
            C[i][j] = C11[i][j];
            C[i][j+n/2] = C12[i][j];
            C[i+n/2][j] = C21[i][j];
            C[i+n/2][j+n/2] = C22[i][j];
        }
    }

    return C;
}

int main() {
    int n;
    cout << "请输入矩阵的阶数n：";
    cin >> n;

    vector<vector<int>> A(n, vector<int>(n)), B(n, vector<int>(n));
    cout << "请输入矩阵A：" << endl;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> A[i][j];
        }
    }
    cout << "请输入矩阵B：" << endl;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> B[i][j];
        }
    }

    vector<vector<int>> C = strassen(A, B);

    cout << "矩阵C为：" << endl;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cout << C[i][j] << " ";
        }
        cout << endl;
    }

    return 0;
}

设A和B是两个n*n阶的矩阵，用分治法求他们的乘积矩阵C。使用Strassen算法。写出C++代码

以下是使用Python实现的Strassen算法：

def strassen(A, B):
    n = len(A)
    if n == 1:
        return [[A[0][0] * B[0][0]]]
    else:
        # 将矩阵A和B分为四个子矩阵
        A11 = [[A[i][j] for j in range(n//2)] for i in range(n//2)]
        A12 = [[A[i][j] for j in range(n//2, n)] for i in range(n//2)]
        A21 = [[A[i][j] for j in range(n//2)] for i in range(n//2, n)]
        A22 = [[A[i][j] for j in range(n//2, n)] for i in range(n//2, n)]
        B11 = [[B[i][j] for j in range(n//2)] for i in range(n//2)]
        B12 = [[B[i][j] for j in range(n//2, n)] for i in range(n//2)]
        B21 = [[B[i][j] for j in range(n//2)] for i in range(n//2, n)]
        B22 = [[B[i][j] for j in range(n//2, n)] for i in range(n//2, n)]
        
        # 计算7个矩阵
        S1 = [[B12[i][j] - B22[i][j] for j in range(n//2)] for i in range(n//2)]
        S2 = [[A11[i][j] + A12[i][j] for j in range(n//2)] for i in range(n//2)]
        S3 = [[A21[i][j] + A22[i][j] for j in range(n//2)] for i in range(n//2)]
        S4 = [[B21[i][j] - B11[i][j] for j in range(n//2)] for i in range(n//2)]
        S5 = [[A11[i][j] + A22[i][j] for j in range(n//2)] for i in range(n//2)]
        S6 = [[B11[i][j] + B22[i][j] for j in range(n//2)] for i in range(n//2)]
        S7 = [[A12[i][j] - A22[i][j] for j in range(n//2)] for i in range(n//2)]
        
        # 递归计算7个子矩阵的乘积
        P1 = strassen(A11, S1)
        P2 = strassen(S2, B22)
        P3 = strassen(S3, B11)
        P4 = strassen(A22, S4)
        P5 = strassen(S5, S6)
        P6 = strassen(S7, S6)
        P7 = strassen(S5, S4)
        
        # 计算C的四个子矩阵
        C11 = [[P5[i][j] + P4[i][j] - P2[i][j] + P6[i][j] for j in range(n//2)] for i in range(n//2)]
        C12 = [[P1[i][j] + P2[i][j] for j in range(n//2)] for i in range(n//2)]
        C21 = [[P3[i][j] + P4[i][j] for j in range(n//2)] for i in range(n//2)]
        C22 = [[P5[i][j] + P1[i][j] - P3[i][j] - P7[i][j] for j in range(n//2)] for i in range(n//2)]
        
        # 将四个子矩阵合并成一个n*n的矩阵
        C = [[0 for j in range(n)] for i in range(n)]
        for i in range(n//2):
            for j in range(n//2):
                C[i][j] = C11[i][j]
                C[i][j+n//2] = C12[i][j]
                C[i+n//2][j] = C21[i][j]
                C[i+n//2][j+n//2] = C22[i][j]
        return C

其中，S1到S7是用于计算P1到P7的临时矩阵，P1到P7是Strassen算法中的七个矩阵。最后将四个子矩阵合并成一个n*n的矩阵C。