Message='itertools.permutations' object is not subscriptable Source=E:\1daer\mmxdzy\1\dzy1\dzy1.py StackTrace: File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 55, in guess_key1 print(next(mp[i][j]), end=" ") ~~~~~^^^ File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 79, in <module> (Current frame) print(guess_key1(cipher_text, words)) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ TypeError: 'itertools.permutations' object is not subscriptable

这个错误是因为您在使用 `next(mp[i][j])` 时，将 `mp[i][j]` 当做了一个可索引的对象，但实际上 `mp[i][j]` 是一个 `itertools.permutations` 对象，不支持索引操作。

如果您想获取 `itertools.permutations` 中的某个元素，可以将其转换为列表或元组，然后再进行索引操作。例如，您可以将 `next(mp[i][j])` 改为 `list(mp[i][j])[0]`，这样就可以获取排列中的第一个元素。

以下是修正后的示例代码：

import itertools

def guess_key1(cipher_text, words):
    # 生成字符组合的排列
    perms = itertools.permutations('ABCDEFGHIJKLMNOPQRSTUVWXYZ', 5)
    for perm in perms:
        # 将排列转换为字典
        key = {chr(i+65): perm[i] for i in range(26)}
        # 解密密文
        plain_text = decrypt(cipher_text, key)
        # 如果解密后的明文包含给定单词，则返回密钥
        if contains_words(plain_text, words):
            return key

def contains_words(text, words):
    # 判断文本中是否包含给定单词
    for word in words:
        if word in text:
            return True
    return False

def decrypt(cipher_text, key):
    # 解密密文
    plain_text = ''
    for ch in cipher_text:
        if ch.isalpha():
            plain_text += key[ch]
        else:
            plain_text += ch
    return plain_text

cipher_text = 'LXYXKXOXRKXYXKXXKLYXKXXYXKXOXRKXYXKXXKLRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKX'
words = ['CRYPTO', 'GOLD', 'SILVER']

# 获取排列器
perms = itertools.permutations('ABCDEFGHIJKLMNOPQRSTUVWXYZ', 5)

# 将排列器转换为二维列表
mp = [[0] * 5 for i in range(5)]
for i in range(5):
    for j in range(5):
        mp[i][j] = perms

# 遍历排列器中的每一个排列
for i in range(5):
    for j in range(5):
        # 获取排列器中的下一个排列
        perm = next(mp[i][j])
        # 输出排列中的第一个元素
        print(list(perm)[0], end="  ")
    print()

在上面的代码中，我们使用 `list(perm)[0]` 来获取排列中的第一个元素，然后输出到控制台上。