Author:
VladMetz

Language:
C/C++

Upload:
10:34 AM 02/01/2022

Download

Platform
Windows

Difficulty:
3.0

Quality:
3.8

Arch:
x86-64

Description

I think it would be interesting to solve my creation, it would be good if you write the keygen in any language The main rule is not to change the principle of the program

• Comments
• Solutions

Sir_Zed on 5:42 PM 02/15/2022: Well i think ur algorithm is broken :) import random,string print("12345678" + (random.choice(string.ascii_letters) * 3)) Res : 12345678ppp 12345678ccc 12345678(any random character) .... ...

ThePhilosopher on 5:18 PM 07/04/2022: It's not broken, the prefix you chose "12345678" can be a working key if the characters at index 9 and 10 can confirm with the algorithm, for example the key "12345678jmg" is not a working key.

Yurri on 11:27 AM 07/11/2022: very easy , I do bruteforce and password : duuhyjjvaee

Marshad on 11:13 AM 07/22/2022: Each string in Flag.txt is a key file = open('Flag.txt','w') for a in range(33,126): for b in range(33,126): for c in range(33,126): for d in range(33,126): for e in range(33,126): for f in range(33,126): if ((e-f)*(e-f)+136*(c-b)*(d-a) == 1632): temp=chr(a) temp+='xx' temp+=chr(b) temp+='x' temp+=chr(c) temp+=chr(d) temp+='xx' temp+=chr(e) temp+=chr(f) file.writelines(temp) file.write("\n") file.close()

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