2020 Zh3r0 部分WP

测试文件：https://wwa.lanzous.com/iEIHddtrkwb

 

crypto

RSA-Warmup

 

就是一个简单的RSA解密

import gmpy2
import rsa
from Crypto.Util.number import long_to_bytes


e = 65537
n = 410498917537920303260266719109576967260993991665256953311742364884150210432915310835865207111036400830428810414329739908868490183839915546875004813875948578775268863603721314519816891107904905552051724797372408241008165216310668239217161510001533176220623603332300737298195416759316147977982808952744735185499042992851
p = 2314713113
q = 177343323987952153300073657598697401625250131662971640715694327551921640853863910709121734305842851476699959583427538092867964922545421095074171644819272503415099832116310007062122212597455875540310089735674473729482935372436656792386332115621247260900026535310987689254169536801391139817672432632365156167627
enc = 281514135280131449141829505417513079250325933715045609473648447005814096770465720188289516151363972033489725088974133391622010638491163251851962623645431655330200732754796040933303093820956529388384535167167317814353993583657429669275504208675670077899340221038968938928780460438883730711177453290473326764507911871504


phin = (q-1)*(p-1)
d = gmpy2.invert(e, phin)

flag = pow(enc, d, n)
flag = long_to_bytes(flag)
print(flag)

 

zh3r0{RSA_1s_Fun}

 

Misc

Welcome to Phase 1

 

zh3r0{is_this_a_real_flag?}

 

Reverse

snakes everywhere
Python 字节码的题，不过相较于上次虎符要简单很多，还是可以利用dis.dis对照翻译，找到关键处的字节码。

 23         216 LOAD_NAME                8 (ciphertext)
            218 LOAD_NAME                7 (xor)
            220 LOAD_NAME                4 (key)
            222 LOAD_NAME               10 (i)
            224 LOAD_CONST              10 (16)
            226 BINARY_MODULO
            228 BINARY_SUBSCR
            230 LOAD_NAME                2 (flag)
            232 LOAD_NAME               10 (i)
            234 BINARY_SUBSCR
            236 CALL_FUNCTION            2
            238 INPLACE_ADD
            240 STORE_NAME               8 (ciphertext)
            242 JUMP_ABSOLUTE          212

翻译为对应的python代码就是

for i in range(16):
    ciphertext[i]  = ciphertext[i] ^ key[i]

最后将ciphertext写入到snake中，因此只需要反向解密就行

 

key = "I_l0v3_r3v3r51ng"
flag = ''
enc = ''
with open(r'C:\Users\10245\Desktop\snake.txt','r', encoding="utf-8") as f:
    enc = f.read()
print (enc)

for i in range(len(enc)):
    flag += chr(ord(enc[i]) ^ ord(key[i%len(key)]))
print (flag)

 

合理猜测，得到flag

zh3r0{Python_disass3mbly_is v3ry_E4sy}