攻防世界-Crypto-easychallenge

题目描述：将文件下载下来，只有一个pyc文件

1. 思路分析

先向chatgpt问下什么是pyc文件：

OK，这里简单总结下：

1. pyc文件是python源码编译后的生成的二进制文件

2. 通过一些库可以逆向出pyc的源代码

那么我们需要做的就是先将源代码还原，还原后再根据具体代码实现找出flag

2. 解题过程

2.1 先逆向出pyc的源代码

我们使用uncompyle6试试（先pip install uncompyle6安装该工具）

然后执行命令：uncompyle6 "42aa1a89e3ae48c38e8b713051557020.pyc" > source.py

这样我们将源代码输出到了source.py中，源代码如下：

# uncompyle6 version 3.9.0
# Python bytecode version base 2.7 (62211)
# Decompiled from: Python 3.10.8 (main, Nov  4 2022, 09:21:25) [GCC 12.2.0]
# Embedded file name: ans.py
# Compiled at: 2018-08-08 23:29:44
import base64

def encode1(ans):
    s = ''
    for i in ans:
        x = ord(i) ^ 36
        x = x + 25
        s += chr(x)

    return s


def encode2(ans):
    s = ''
    for i in ans:
        x = ord(i) + 36
        x = x ^ 36
        s += chr(x)

    return s


def encode3(ans):
    return base64.b32encode(ans)


flag = ' '
print 'Please Input your flag:'
flag = raw_input()
final = 'UC7KOWVXWVNKNIC2XCXKHKK2W5NLBKNOUOSK3LNNVWW3E==='
if encode3(encode2(encode1(flag))) == final:
    print 'correct'
else:
    print 'wrong'
# okay decompiling 42aa1a89e3ae48c38e8b713051557020.pyc

2.2 解码出flag

从代码中分析，代码对flag进行了三层编码，那么我们需要对这三层编码一一进行解码，我们按照编码的顺序反着进行解码即可，调整下代码如下：

import base64

def decode1(ans):
    s = ''
    for i in ans:
        # x = ord(i) ^ 36
        x = ord(i) - 25
        x = x ^ 36
        s += chr(x)

    return s


def decode2(ans):
    s = ''
    for i in ans:
        # x = ord(i) + 36
        x = i ^ 36
        x = x - 36
        s += chr(x)

    return s


def decode3(ans):
    return base64.b32decode(ans)


#flag = ' '
#print 'Please Input your flag:'
#flag = raw_input()
final = 'UC7KOWVXWVNKNIC2XCXKHKK2W5NLBKNOUOSK3LNNVWW3E==='
flag = decode1(decode2(decode3(final)))
print(flag)

执行该脚本获取flag即可：cyberpeace{interestinghhhhh}