strassen矩阵乘法的串行程序，并输出运行时间

以下是串行实现strassen矩阵乘法的示例代码，并输出运行时间：

import time

# 定义矩阵乘法函数
def matrix_multiply(a, b):
    ra, ca = a.shape  # 矩阵a的行数和列数
    rb, cb = b.shape  # 矩阵b的行数和列数

    # 判断能否相乘
    if ca != rb:
        raise ValueError("Matrix shapes are not aligned. %s != %s" % (ca, rb))

    c = np.zeros((ra, cb))  # 结果矩阵c初始化

    # 矩阵相乘
    for i in range(ra):
        for j in range(cb):
            for k in range(ca):
                c[i, j] += a[i, k] * b[k, j]

    return c

# 定义strassen矩阵乘法函数
def strassen_mult(a, b):
    # 如果矩阵尺寸小于等于64，则采用暴力矩阵乘法
    if a.shape[0] <= 64:
        return matrix_multiply(a, b)

    # 矩阵尺寸一半
    m = a.shape[0] // 2

    # 划分为子问题
    a11, a12 = a[:m, :m], a[:m, m:]
    a21, a22 = a[m:, :m], a[m:, m:]
    b11, b12 = b[:m, :m], b[:m, m:]
    b21, b22 = b[m:, :m], b[m:, m:]

    # 7个子问题
    p1 = strassen_mult(a11 + a22, b11 + b22)
    p2 = strassen_mult(a21 + a22, b11)
    p3 = strassen_mult(a11, b12 - b22)
    p4 = strassen_mult(a22, b21 - b11)
    p5 = strassen_mult(a11 + a12, b22)
    p6 = strassen_mult(a21 - a11, b11 + b12)
    p7 = strassen_mult(a12 - a22, b21 + b22)

    # 计算结果矩阵的四个子矩阵
    c11 = p1 + p4 - p5 + p7
    c12 = p3 + p5
    c21 = p2 + p4
    c22 = p1 - p2 + p3 + p6

    # 合并子矩阵
    c = np.zeros((a.shape[0], b.shape[1]))
    c[:m, :m], c[:m, m:], c[m:, :m], c[m:, m:] = c11, c12, c21, c22

    return c

# 生成待计算的矩阵a和b
a = np.random.rand(2048, 2048)
b = np.random.rand(2048, 2048)

# 记录运行时间
start_time = time.time()

# 调用strassen矩阵乘法函数
c = strassen_mult(a, b)

end_time = time.time()

# 输出运行时间
print("运行时间：", end_time - start_time, "秒")

注意：以上代码为Python代码，需要结合相关库使用。运行时间与测试环境有关，仅供参考。