NSSCTF_RE_简简单单的解密

import base64,urllib.parse
key = "HereIsFlagggg"
flag = "xxxxxxxxxxxxxxxxxxx"

s_box = list(range(256))
j = 0
for i in range(256):
    j = (j + s_box[i] + ord(key[i % len(key)])) % 256
    s_box[i], s_box[j] = s_box[j], s_box[i]
res = []
i = j = 0
for s in flag:
    i = (i + 1) % 256
    j = (j + s_box[i]) % 256
    s_box[i], s_box[j] = s_box[j], s_box[i]
    t = (s_box[i] + s_box[j]) % 256
    k = s_box[t]
    res.append(chr(ord(s) ^ k))
cipher = "".join(res)
crypt = (str(base64.b64encode(cipher.encode('utf-8')), 'utf-8'))
enc = str(base64.b64decode(crypt),'utf-8')
enc = urllib.parse.quote(enc)
print(enc)
# enc = %C2%A6n%C2%87Y%1Ag%3F%C2%A01.%C2%9C%C3%B7%C3%8A%02%C3%80%C2%92W%C3%8C%C3%BA

参照大佬的脚本:

 

 

import base64,urllib.parse
flag=''
enc="%C2%A6n%C2%87Y%1Ag%3F%C2%A01.%C2%9C%C3%B7%C3%8A%02%C3%80%C2%92W%C3%8C%C3%BA"
key = "HereIsFlagggg"
crypt=urllib.parse.unquote(enc)
s_box = list(range(256))
j = 0
for i in range(256):
    j = (j + s_box[i] + ord(key[i % len(key)])) % 256
    s_box[i], s_box[j] = s_box[j], s_box[i]
i = j = 0
for s in crypt:
    i = (i + 1) % 256
    j = (j + s_box[i]) % 256
    s_box[i], s_box[j] = s_box[j], s_box[i]
    t = (s_box[i] + s_box[j]) % 256
    k = s_box[t]
    flag+=chr(ord(s) ^ k)
print(flag)

当初没咋看懂这个,听说是RC4,

百度词条又这么一句话

 

 

 

 所以直接抄了源代码的加密代码,然后quote的那个加密,:

NSSCTF{REAL_EZ_RC4}

 

 

 直接unquote解密就好了

博客地址:Python3 urllib.parse 常用函数示例 - <Hbw> - 博客园 (cnblogs.com)

最后得到flag:NSSCTF{REAL_EZ_RC4}