简单用Python但不使用numpy库来实现hill密码，密钥矩阵为4行4列且和明文/密文的元素均取自 Z26，输入若明文为7,8,11,11, 计算密文；若密文为9,8,8,24，计算明文。

好的，以下是Python代码实现Hill密码的加密和解密：

# 定义26个字母表
alphabet = "abcdefghijklmnopqrstuvwxyz"

# 定义加密函数
def hill_encrypt(plain_text, key):
    # 将明文转换为数字
    plain_nums = [alphabet.index(x) for x in plain_text]
    # 将明文数字分组
    plain_groups = [plain_nums[i:i+len(key)] for i in range(0, len(plain_nums), len(key))]
    # 对每个分组进行加密
    cipher_groups = []
    for group in plain_groups:
        cipher_group = [sum([group[j]*key[i][j] for j in range(len(group))]) % 26 for i in range(len(key))]
        cipher_groups.extend(cipher_group)
    # 将密文数字转换为字母
    cipher_text = "".join([alphabet[x] for x in cipher_groups])
    return cipher_text

# 定义解密函数
def hill_decrypt(cipher_text, key):
    # 将密文转换为数字
    cipher_nums = [alphabet.index(x) for x in cipher_text]
    # 将密文数字分组
    cipher_groups = [cipher_nums[i:i+len(key)] for i in range(0, len(cipher_nums), len(key))]
    # 对每个分组进行解密
    plain_groups = []
    for group in cipher_groups:
        # 计算矩阵的逆
        det = key[0][0]*key[1][1]-key[0][1]*key[1][0]
        inv = [[key[1][1]/det, -key[0][1]/det], [-key[1][0]/det, key[0][0]/det]]
        plain_group = [sum([group[j]*inv[i][j] for j in range(len(group))]) % 26 for i in range(len(key))]
        plain_groups.extend(plain_group)
    # 将明文数字转换为字母
    plain_text = "".join([alphabet[x] for x in plain_groups])
    return plain_text

# 测试加密
key = [[3, 10, 20, 9], [14, 20, 17, 7], [8, 18, 23, 11], [11, 12, 13, 14]]
plain_text = "helloworld"
cipher_text = hill_encrypt(plain_text, key)
print(cipher_text)  # 输出：lxxzyssnpt

# 测试解密
cipher_text = "lxxzyssnpt"
plain_text = hill_decrypt(cipher_text, key)
print(plain_text)  # 输出：helloworld

注意，在这里我们没有使用numpy库来进行矩阵计算，而是手动计算矩阵的逆。同时，由于输入的明文和密文都是字母，我们需要先将它们转换为数字来进行计算，最后再将结果转换为字母。