0x00 简介

培根密码，培根所用的密码是一种本质上用二进制数设计的，没有用通常的0和1来表示，而是采用a和b 培根密码加密有两种方式: A aaaaa B aaaab C aaaba D aaabb E aabaa F aabab G aabba H aabbb I abaaa J abaab K ababa L ababb M abbaa N abbab O abbba P abbbb Q baaaa R baaab S baaba T baabb U babaa V babab W babba X babbb Y bbaaa Z bbaab 方式二： a AAAAA g AABBA n ABBAA t BAABA b AAAAB h AABBB o ABBAB u-v BAABB c AAABA i-j ABAAA p ABBBA w BABAA d AAABB k ABAAB q ABBBB x BABAB e AABAA l ABABA r BAAAA y BABBA f AABAB m ABABB s BAAAB z BABBB

通常加密者只需要两种不同的字体或使用大小写来代替ab即可，例如明文为bling，加密为：aaaab ababb abaaa abbab aabba，此时再随意找句句子，使用大小写来代替ab。密文如下：good GoOd STuDy day DAy Up hAHa

0x01 解密代码

#! python3 # -*- coding: utf-8 -*- # @Time : 2017/8/26 上午12:07 # @Author : BlingBling # @File : Baconian.py # @Software: PyCharm Community Edition import re class Baconian(): alphabet = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'] first_cipher = ["aaaaa", "aaaab", "aaaba", "aaabb", "aabaa", "aabab", "aabba", "aabbb", "abaaa", "abaab", "ababa", "ababb", "abbaa", "abbab", "abbba", "abbbb", "baaaa", "baaab", "baaba", "baabb", "babaa", "babab", "babba", "babbb", "bbaaa", "bbaab"] second_cipher = ["aaaaa", "aaaab", "aaaba", "aaabb", "aabaa", "aabab", "aabba", "aabbb", "abaaa", "abaaa", "abaab", "ababa", "ababb", "abbaa", "abbab", "abbba", "abbbb", "baaaa", "baaab", "baaba", "baabb", "baabb", "babaa", "babab", "babba", "babbb"] def __init__(self, str): self.str = str def decode(self): str = self.str.lower() str_array = re.findall(".{5}", str) decode_str1 = "" decode_str2 = "" for key in str_array: for i in range(0,26): if key == Baconian.first_cipher[i]: decode_str1 += Baconian.alphabet[i] if key == Baconian.second_cipher[i]: decode_str2 += Baconian.alphabet[i] print(decode_str1) print(decode_str2) if __name__ == '__main__': str = input("please input string to decode:\n") bacon = Baconian(str) bacon.decode()

总结：

培根密码：base64解码后直接培根解密http://ctf.ssleye.com/baconian.html Crypto2：RSA小明文攻击，得出输出也就是C后，使用https://xz.aliyun.com/t/6459#toc-34里面的脚本。

MISC2：16进制编辑器打开，逆序答案在内。

Re1 ida 打开搜索字符串 flag，一堆 flag 中间只有一个有引用的，就是正确的。