letters = ['7', 'u', 'b', 'u', 'r', 'l', 'n', ' ', '5', 'o', 'j', 'c', '3', '7', '1', 'x', 'd', 'x', 'm', '8', 'a', 'v', 'f', 'd', 'z', 'z', 'd', 'l', 'v', 'o', 'q', 'j', 'u', '2', 'o', 'l', 'v', '6', 'i', 'o', 'i', '5', '9', 'b', 'c', 'i', 's', 'a', 'i', '2', '!', '8', 's', 't', '9', 'r', 'x', 'w', 'j', '1', '5', '5', '3', 'm', '9', '6', 'r', 'w', '9', 'd', 'd', 'r', 'y', '3', 'p', 'h', 'f', '9', '1', 'b', 'w', 'u', 'c', 'e', 'r', '3', 'i', 'i', 'x', '7', 'x', '2', 'n', 'p', 'a', '4', 'b', 'f', 'w', ' '] letters.append('!'） answer = letters answer.append(str(letters.count('x'))) ​answer.append(letters[-1]) answer.extend(letters[-3:]) print(''.join(answer))

你想问的是这段 Python 代码会输出什么结果，对吧？答案是：uburln 5ojc371xdxmm8avfdzzdlvoqju2olv6ioi59bcisai2!wuceri7x2npa4bfw!5

解释如下：代码首先定义了一个包含多个字符的列表 letters，并且通过 append 方法添加了一个 "!"。然后，代码接着对 letters 进行了一系列操作，包括添加元素、计算元素数量、获取元素等等。最后，代码通过 join 方法将结果输出为一个字符串。

相关问题

def guess_key2(cipher_text, word1, word2 ,word3): letter_frequency = get_letter_frequency(cipher_text.lower()) excluded_letters = [letter for letter in letter_frequency.keys() if letter_frequency[letter] == 0] sorted_letters = sorted([letter for letter in letter_frequency.keys() if letter_frequency[letter] > 0], key=lambda x: letter_frequency[x], reverse=True) print("Excluded letters:", excluded_letters) print() most_common_letters_m = [sorted_letters[:1], sorted_letters[1:7], sorted_letters[7:10], sorted_letters[10:15], sorted_letters[15:20], sorted_letters[20:26]] # high 8 middle 8 low 10 f1 = ['e'] f2 = ['a', 'i', 'r', 't', 'o', 'n'] f3 = ['s', 'l', 'c'] f4 = ['u', 'p', 'm', 'd', 'h'] f5 = ['g', 'b', 'y', 'f', 'v'] mf = [f1, f2, f3, f4, f5] mp = [[j for j in range(len(mf[i]))] for i in range(len(mf))] flag = True while(flag): key = {} row_permutations = [itertools.permutations(row) for row in mp] matrix_permutations = itertools.product(*row_permutations) for permutation in matrix_permutations: for i in range(len(mp)): for j in range(len(mp[i])): key[most_common_letters_m[i][permutation[i][j]]] = mf[i][j] decrypted_text = decrypt(cipher_text, key) if is_plaintext(decrypted_text, word1, word2 ,word3): flag=False return key 这里每个key不包含全部的26组对应关系，请修改以将他们补齐

可以将代码修改为：

import itertools
import random

def guess_key2(cipher_text, word1, word2, word3):
    letter_frequency = get_letter_frequency(cipher_text.lower())
    excluded_letters = [letter for letter in letter_frequency.keys() if letter_frequency[letter] == 0]
    sorted_letters = sorted([letter for letter in letter_frequency.keys() if letter_frequency[letter] > 0], key=lambda x: letter_frequency[x], reverse=True)
    print("Excluded letters:", excluded_letters)
    print()
    most_common_letters_m = [sorted_letters[:1], sorted_letters[1:7], sorted_letters[7:10], sorted_letters[10:15], sorted_letters[15:20], sorted_letters[20:26]]
    # high 8 middle 8 low 10
    f1 = ['e']
    f2 = ['a', 'i', 'r', 't', 'o', 'n']
    f3 = ['s', 'l', 'c']
    f4 = ['u', 'p', 'm', 'd', 'h']
    f5 = ['g', 'b', 'y', 'f', 'v']
    mf = [f1, f2, f3, f4, f5]
    mp = [[j for j in range(len(mf[i]))] for i in range(len(mf))]
    flag = True
    while flag:
        key = {}
        row_permutations = [itertools.permutations(row) if len(row) > 1 else row for row in mp]
        matrix_permutations = itertools.product(*row_permutations)
        for permutation in matrix_permutations:
            for i in range(len(permutation)):
                if len(permutation[i]) == 1:
                    key[most_common_letters_m[i][0]] = mf[i][permutation[i][0]]
                else:
                    for j in range(len(permutation[i])):
                        key[most_common_letters_m[i][permutation[i][j]]] = mf[i][j]
            # 补齐未配对的字母
            remaining_letters = set('abcdefghijklmnopqrstuvwxyz') - set(key.keys())
            while len(remaining_letters) > 0:
                letter = random.sample(remaining_letters, 1)[0]
                value = random.sample(set('abcdefghijklmnopqrstuvwxyz') - set(key.values()), 1)[0]
                key[letter] = value
                remaining_letters = set('abcdefghijklmnopqrstuvwxyz') - set(key.keys())
            decrypted_text = decrypt(cipher_text, key)
            if is_plaintext(decrypted_text, word1, word2, word3):
                flag = False
                break
    return key

这样，我们在生成 key 的过程中，对于每个位置，如果对应的单词只包含一个字母，则直接将该字母和对应的字母配对，否则使用 permutations() 函数生成所有可能的排列，然后将排列中的字母按照顺序与 mf 中的字母配对。然后，我们使用随机字母补齐未配对的字母，直到 key 包含 26 对对应关系。最后，我们检查解密后的明文是否满足给定的单词列表，如果满足，则返回 key。

需要注意的是，这里的随机字母选择算法并不完美，如果 mf 中的字母和已配对的字母集合中的字母的交集不为空，可能会导致算法产生死循环。如果需要更安全和更合理的随机字母选择算法，可以参考 Python 标准库中的 secrets 模块。

guess_key2(cipher_text, word1, word2 ,word3): letter_frequency = get_letter_frequency(cipher_text.lower()) excluded_letters = [letter for letter in letter_frequency.keys() if letter_frequency[letter] == 0] sorted_letters = sorted([letter for letter in letter_frequency.keys() if letter_frequency[letter] > 0], key=lambda x: letter_frequency[x], reverse=True) print("Excluded letters:", excluded_letters) print() #if len(excluded_letters)<5 most_common_letters_m = [sorted_letters[15:21] ,sorted_letters[10:15] ,sorted_letters[7:10], sorted_letters[1:7]] #反向 #f1 = ['e'] f2 = ['a', 'i', 'r', 't', 'o', 'n'] f3 = ['s', 'l', 'c'] f4 = ['u', 'p', 'm', 'd', 'h'] f5 = ['g', 'b', 'y', 'f', 'v', 'w'] f6 = ['k', 'x', 'z', 'q', 'j'] mf = [f5, f4, f3, f2] mp = [[j for j in range(len(mf[i]))] for i in range(len(mf))] flag = True while(flag): key = { sorted_letters[0]: 'e'} row_permutations = [itertools.permutations(row) for row in mp] matrix_permutations = itertools.product(*row_permutations) for permutation in matrix_permutations: for i in range(len(mp)): for j in range(len(mp[i])): key[most_common_letters_m[i][permutation[i][j]]] = mf[i][j] if len(key) < len (sorted_letters): for i in range (len(sorted_letters)-len(key)): key.update({sorted_letters[len(key)+i-1]:f6[i]}) decrypted_text = decrypt(cipher_text, key) if is_plaintext(decrypted_text, word1, word2 ,word3): flag=False return key时间复杂度是否过高？

这段代码的时间复杂度非常高，因为它使用了多重嵌套的循环和迭代器，这会导致执行时间随着输入数据的增加而指数级增加。具体来说，这段代码的时间复杂度是O(N^M)，其中N是字母表中字母的数量，M是最常见的字母组合的数量。如果输入的密文很长，这个时间复杂度可能会变得非常大。因此，这段代码应该尽可能地避免重复计算，以及尝试减少嵌套循环的数量。