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;
}