惜字如金_2.0.md

惜字如金 2.0

比去年的简单。关键就是恢复 cod_dict，根据后面的 check_equals 可知 cod_dict 每个字符串长度原本应该是 24，惜字如金化后长度变成了23。

直接穷举所有可能的原始字符串，看哪个组合能出 flag 就好：

import re
from itertools import product


def get_candidates(s):
    vowel_set = 'AEIOUaeiou'
    parts = re.split(r'([a-zA-Z]+)', s)

    for i in range(len(parts)):
        if parts[i].isalpha():
            left_s = "".join(parts[:i])
            right_s = "".join(parts[i+1:])
            word = parts[i]

            for j in range(len(word)):
                if word[j] not in vowel_set:
                    yield left_s + word[:j] + word[j] + word[j:] + right_s

            if word[-1] not in vowel_set:
                yield left_s + word + 'e' + right_s


t = []
t += ['nymeh1niwemflcir}echaet']
t += ['a3g7}kidgojernoetlsup?h']
t += ['ulw!f5soadrhwnrsnstnoeq']
t += ['ct{l-findiehaai{oveatas']
t += ['ty9kxborszstguyd?!blm-p']
cod_dict_full = []

for s in t:
    cod_dict_full.append(list(get_candidates(s)))

input_codes = [53, 41, 85, 109, 75, 1, 33, 48, 77, 90,
               17, 118, 36, 25, 13, 89, 90, 3, 63, 25,
               31, 77, 27, 60, 3, 118, 24, 62, 54, 61,
               25, 63, 77, 36, 5, 32, 60, 67, 113, 28]

for cod_dict in product(*cod_dict_full):
    cod_dict_str = ''.join(cod_dict)
    output_chars = [cod_dict_str[c] for c in input_codes]
    flag = ''.join(output_chars)
    try:
        if flag.index('flag{') == 0 and flag.index('}') == len(flag) - 1:
            print(flag)
    except ValueError:
        pass