Bugku 加密 python writeup

一上来就给了两个文件，一个是加密的源代码，一个是加密过程文件，

challenge.py

                                                                N1ES.py

N1ES.py里一共有四个函数，一个类，类里含有两个函数，除了最后一个encrypt函数外其他函数都是在对key进行运算，然后通过key来对flag进行加密，所以我直接跑了一下程序，获得了key加密后的数据，然后只对encrypt函数进行逆向
解密脚本：

Kn=[['~', 'w', 'Y', 'k', 'k', '\x02', '\x05', '\x05'],['w', 'd', '}', '\x14', '?', '\x13', '\x04', 'W'],['l', '6', '\x08', '\x04', '\x13', '3', '\x19', '\x10'],['\x08', 'P', '2', '\x02', '/', 'W', '/', 'W'],['\x08', '\x14', '?', '@', 'W', '^', ' ', 'k'],['\x1b', '6', '^', '(', 'M', 'Y', '\x19', '\x02'],['3', 'f', 'w', '(', '\x13', '}', '\x08', 'u'],['=', '_', '\x13', 'M', '2', '=', '@', '\x04'],['z', '_', '~', '\x08', 'L', 'f', '\x19', 'z'],['I', 'Y', '\x01', '}', '/', '}', 'L', 'o'],['\x19', '\x05', '3', '\x01', 'z', 'w', '~', '?'],['L', 'B', '~', '\x13', '@', '6', '@', '\x05'],['\x08', 'd', '\x13', 'L', '^', '?', 'L', 'u'],['\x05', '{', 'M', 'P', 'M', '\n', 'z', 'P'],['k', '~', 'k', '/', 'o', 'u', '\x19', '\x04'],['o', 'k', '(', '\x13', 'I', 'f', ' ', '='],['~', '\x04', '\x08', '^', '\x02', '\n', '6', '3'],['/', '\x05', 'w', '2', ' ', 'd', '\x13', '6'],[' ', '/', '}', '?', '\x04', '}', 'z', '\x19'],['\x05', '\n', '\n', 'l', '\x02', 'l', '^', 'l'],['k', '3', '}', '\x19', 'u', 'I', ' ', '^'],['~', 'B', '\x02', '}', 'k', '\x05', '\x02', '/'],['\n', '\x05', '^', '^', 'P', '}', '!', '{'],['\x08', 'W', 'u', 'o', ' ', '2', 'd', '\x04'],['/', 'W', 'w', '\x08', 'z', '\x19', '@', 'I'],['\x14', ' ', 'P', '!', '6', '6', ' ', '}'],['(', '!', '\x01', '\x08', 'd', '\x08', 'w', '?'],['u', 'W', '@', '\x13', '}', '~', '6', 'o'],['3', 'B', 'd', '\x01', 'W', '2', '\n', '6'],['}', '\x08', '6', '\x19', '&', '\x04', 'k', 'u'],['\x13', '2', '2', '(', '\x19', '{', '/', 'w'],['\x02', 'Y', ' ', 'W', '\x08', 'u', '\x01', 'I']]

import base64

s=base64.b64decode('HRlgC2ReHW1/WRk2DikfNBo1dl1XZBJrRR9qECMNOjNHDktBJSxcI1hZIz07YjVx')

flag=[]

for i in range(3):

    flag.append(s[i*16:(i+1)*16])

from z3 import *

def fun(a,b):

    x=[BitVec('x%d'%i,32) for i in range(8)]

    solver=Solver()

    res=''

    for i in range(len(a)):

        exec("solver.add(x[i]-2*(x[i]&ord(b[i]))==ord(a[i])-ord(b[i]))")

        solver.check()

        try:

            exec("res+=chr(solver.model()[x[i]].as_long())")

        except:

            print solver

    return res

res=''

for i in flag:

    L=i[:8]

    R=i[8:]

    L,R=R,L

    for k in range(32):

        L,R=R,fun(L,Kn[k])

    res+=L+R

print res