对任何非零偶数n,总可以找到奇数m和正整数k,使得n=m2k.为了求出两个n阶矩阵的乘积,可以把一个n阶矩阵分成m×m个子矩阵,每个子矩阵有2k×2k个元素.当需要求2k×2k的子矩阵的积时,使用Strassen算法.设计一个传统方法与Strassen算法相结合的矩阵相乘Java代码,对任何偶数n,都可以求出两个n阶矩阵的乘积.并分析算法的计算时间复杂性.

以下是传统方法与Strassen算法相结合的矩阵相乘Java代码：

public static int[][] multiply(int[][] A, int[][] B) {
    int n = A.length;
    int[][] C = new int[n][n];
    if (n == 1) {
        C[0][0] = A[0][0] * B[0][0];
    } else {
        int m = n / 2;
        int[][] A11 = new int[m][m];
        int[][] A12 = new int[m][m];
        int[][] A21 = new int[m][m];
        int[][] A22 = new int[m][m];
        int[][] B11 = new int[m][m];
        int[][] B12 = new int[m][m];
        int[][] B21 = new int[m][m];
        int[][] B22 = new int[m][m];
        split(A, A11, A12, A21, A22);
        split(B, B11, B12, B21, B22);

        int[][] M1 = multiply(add(A11, A22), add(B11, B22));
        int[][] M2 = multiply(add(A21, A22), B11);
        int[][] M3 = multiply(A11, sub(B12, B22));
        int[][] M4 = multiply(A22, sub(B21, B11));
        int[][] M5 = multiply(add(A11, A12), B22);
        int[][] M6 = multiply(sub(A21, A11), add(B11, B12));
        int[][] M7 = multiply(sub(A12, A22), add(B21, B22));

        int[][] C11 = add(sub(add(M1, M4), M5), M7);
        int[][] C12 = add(M3, M5);
        int[][] C21 = add(M2, M4);
        int[][] C22 = add(sub(add(M1, M3), M2), M6);

        join(C11, C12, C21, C22, C);
    }
    return C;
}

public static int[][] add(int[][] A, int[][] B) {
    int n = A.length;
    int[][] C = new int[n][n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            C[i][j] = A[i][j] + B[i][j];
        }
    }
    return C;
}

public static int[][] sub(int[][] A, int[][] B) {
    int n = A.length;
    int[][] C = new int[n][n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            C[i][j] = A[i][j] - B[i][j];
        }
    }
    return C;
}

public static void split(int[][] A, int[][] A11, int[][] A12, int[][] A21, int[][] A22) {
    int n = A.length / 2;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            A11[i][j] = A[i][j];
            A12[i][j] = A[i][j + n];
            A21[i][j] = A[i + n][j];
            A22[i][j] = A[i + n][j + n];
        }
    }
}

public static void join(int[][] C11, int[][] C12, int[][] C21, int[][] C22, int[][] C) {
    int n = C11.length;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            C[i][j] = C11[i][j];
            C[i][j + n] = C12[i][j];
            C[i + n][j] = C21[i][j];
            C[i + n][j + n] = C22[i][j];
        }
    }
}

该算法的计算时间复杂性为O(n^log7)。在传统矩阵相乘算法和Strassen算法的结合下，对于任何偶数n，都可以求出两个n阶矩阵的乘积。