LINK：https://www.root-me.org/en/Challenges/Cracking/PYC-ByteCode
使用Pyc的逆向脚本uncompyle6进行反编译，该逆向脚本内容如下：
#!/usr/bin/python
# EASY-INSTALL-ENTRY-SCRIPT: 'uncompyle6==3.2.4','console_scripts','uncompyle6'
__requires__ = 'uncompyle6==3.2.4'
import re
import sys
from pkg_resources import load_entry_point
if __name__ == '__main__':
sys.argv[0] = re.sub(r'(-script\.pyw?|\.exe)?$', '', sys.argv[0])
sys.exit(
load_entry_point('uncompyle6==3.2.4', 'console_scripts', 'uncompyle6')()
)
得到反编译结果如下：
# uncompyle6 version 3.2.4
# Python bytecode 3.1 (3151)
# Decompiled from: Python 3.7.2 (default, Jan 10 2019, 23:51:51)
# [GCC 8.2.1 20181127]
# Embedded file name: crackme.py
# Compiled at: 2013-07-02 15:00:05
if __name__ == '__main__':
print('Welcome to the RootMe python crackme')
PASS = input('Enter the Flag: ')
KEY = 'I know, you love decrypting Byte Code !'
I = 5
SOLUCE = [57, 73, 79, 16, 18, 26, 74, 50, 13, 38, 13, 79, 86, 86, 87]
KEYOUT = []
for X in PASS:
KEYOUT.append((ord(X) + I ^ ord(KEY[I])) % 255)
I = (I + 1) % len(KEY)
if SOLUCE == KEYOUT:
print('You Win')
else:
print('Try Again !')
# okay decompiling ch19.pyc
介绍一个源代码涉及到的函数：ord()
描述
ord() 函数是 chr() 函数（对于8位的ASCII字符串）或 unichr() 函数（对于Unicode对象）的配对函数，它以一个字符（长度为1的字符串）作为参数，返回对应的 ASCII 数值，或者 Unicode 数值，如果所给的 Unicode 字符超出了你的 Python 定义范围，则会引发一个 TypeError 的异常。
语法
以下是 ord() 方法的语法:
ord(c)
参数
- c -- 字符。
返回值
返回值是对应的十进制整数。
由源代码可以编写出破解脚本如下：
#!/usr/bin/env python
if __name__ == '__main__':
print('Welcome to the RootMe python crackme')
# PASS = input('Enter the Flag: ')
KEY = 'I know, you love decrypting Byte Code !'
DICT = "abcdefghijklmnopqrstuvwxyz1234567890ABCDEFGHIJKLMNOPQRSTUVWXYZ!@#$%`+^&*(){}[]\";?'\/<>~=:-_. |,"
I = 5
SOLUCE = [57, 73, 79, 16, 18, 26, 74, 50, 13, 38, 13, 79, 86, 86, 87]
KEYOUT = []
RESULT = []
for i in SOLUCE:
for X in DICT:
DATA=(ord(X) + I ^ ord(KEY[I])) % 255
if DATA==i:
KEYOUT.append(DATA)
I = (I + 1) % len(KEY)
RESULT.append(X)
break
print(RESULT)
['I', '_', 'h', 'a', 't', 'e', '_', 'R', 'U', 'B', 'Y', '_', '!', '!', '!']
逆向运算得到password为： I_hate_RUBY_!!!