用Python实现矩阵连乘问题，并在每行后面写出相应的代码

以下是Python实现矩阵连乘问题的代码：

import sys

def matrix_chain_order(p):
    n = len(p) - 1
    m = [[0] * (n + 1) for i in range(n + 1)]
    s = [[0] * (n + 1) for i in range(n + 1)]
    for l in range(2, n + 1):
        for i in range(1, n - l + 2):
            j = i + l - 1
            m[i][j] = sys.maxsize
            for k in range(i, j):
                q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j]
                if q < m[i][j]:
                    m[i][j] = q
                    s[i][j] = k
    return m, s

def print_optimal_parens(s, i, j):
    if i == j:
        print("A" + str(i), end="")
    else:
        print("(", end="")
        print_optimal_parens(s, i, s[i][j])
        print_optimal_parens(s, s[i][j] + 1, j)
        print(")", end="")

p = [30, 35, 15, 5, 10, 20, 25]
m, s = matrix_chain_order(p)
print_optimal_parens(s, 1, len(p) - 1)

该代码实现了动态规划算法来解决矩阵连乘问题，并输出最优的括号化方案。