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PyLu

蟒蛇帕鲁#

文件

解包#

看图标/扔进DIE，使用PyInstaller打包的，开始解包

另外，这里的pycdc并没有给可执行文件，需要手动编译

  • 安装 cmake
  • git clone https://github.com/zrax/pycdc.git
  • cmake . msbuild pycdc.sln
  • 用 VS 打开，编译

已经把编译好的文件扔进[[Python工具和包|Python常用工具]]了

参考

【python逆向 pyc反编译】python逆向全版本通杀-CSDN博客

Decompiling recent python versions (up to 3.12) - Ring 0x00 (idafchev.github.io)

分析#

from Crypto.Util.number import bytes_to_long

def enc(key):
    R = bytes_to_long(b'Welcome To PaluCTF!')
    MOD = 2 ** 418
    R = R ^ R - 60 >> 24
    R = R ^ R - 60 << 88
    R ^= key
    R = -R ** 2 * 2024 % MOD
    R = R * key % MOD
    return R

flag = input('Welcome To PaluCTF!\nInput FLAG:')
m = bytes_to_long(flag.encode())
cor = 0x2E441F765514CCA89173554726494D37E9FBE774B6F807BC5F6E71117530CE3D7DB5F70554C03CD9055F4E42969600904DF1F4DB8L
if enc(m) == cor:
    print('Congratulation!')
    return None
print('Wrong FLAG!')

稍微修改下代码，使得可读性提高一些

def enc(key):
    R = bytes_to_long(b'Welcome To PaluCTF!')
    MOD = 2 ** 418
    R = R ^ ((R - 60) >> 24)
    R = R ^ ((R - 60) << 88)
    # 此时 hex(R) # 0x57656c340a01064f390a00536b201128393569390a00040e4c2522383362

    KeyXor = key ^ R
    temp = (KeyXor ** 2) * (key)
    R = ((-(temp)) * 2024) % MOD
    return R

flag = input()
m = bytes_to_long(flag.encode())
cor = 0x2E441F765514CCA89173554726494D37E9FBE774B6F807BC5F6E71117530CE3D7DB5F70554C03CD9055F4E42969600904DF1F4DB8
if enc(m) == cor:
    print('Congratulation!')
    exit()
print('Wrong FLAG!')

所以算法是

graph TD
  A[输入str] -->|encode| B[bytes];
  B -->|bytes_to_long| K[大数key];
  K --> Z;

  W['Welcome To PaluCTF!'] -->|计算| R;
  R[一个固定的大数R] --> Z;

  
  Z[异或 乘法 取模运算] -->O[一个大数];

以上算法因为带有==异或/平方/取模/乘法==运算，导致很难找到逆运算，

但是另外，这些运算并没有改变各个位的位置，这意味着我们可以从个位开始，逐位爆破

EXP#

from Crypto.Util.number import bytes_to_long
from string import printable

cor = 0x2E441F765514CCA89173554726494D37E9FBE774B6F807BC5F6E71117530CE3D7DB5F70554C03CD9055F4E42969600904DF1F4DB8

def enc(key):
    R = bytes_to_long(b'Welcome To PaluCTF!')
    MOD = 2 ** 418
    R = R ^ R - 60 >> 24
    R = R ^ R - 60 << 88
    R ^= key
    R = -R ** 2 * 2024 % MOD
    R = R * key % MOD
    return R

def find_next(current_str: str):
    print(current_str)
    length = len(current_str)
    if (cor >> (length * 8)) == 0:
        return current_str

    for c in printable:
        s = (c + current_str)
        e = enc(bytes_to_long(s.encode()))
        if (e >> (length * 8)) & 0xFF == cor >> (length * 8) & 0xFF:
            r = find_next(c + current_str)
            if r == None:
                continue
            else:
                return r
    
    return None

find_next("")

EXP2#

看了下第一名的wp，用到了z3

Z3官方文档 | Introduction | Online Z3 Guide (microsoft.github.io)

Z3基础学习（一） 从例子入门

帕鲁杯 Writeup (yuque.com)

import z3
from Crypto.Util.number import *

def enc(key):
    R = bytes_to_long(b"Welcome To PaluCTF!")
    MOD = 2**418
    R = R ^ ((R - 60) >> 24)
    R = R ^ ((R - 60) << 88)
    R ^= key
    R = (-R * R * 2024) % MOD  
    R = (R * key) % MOD
    return R

res = 0x2E441F765514CCA89173554726494D37E9FBE774B6F807BC5F6E71117530CE3D7DB5F70554C03CD9055F4E42969600904DF1F4DB8
solver = z3.Solver()
key = z3.BitVec("key", 418)
solver.add(enc(key) == res)
solver.check()
m = solver.model()
flag = long_to_bytes(m[key].as_long())
print(flag)

太抽象了