VladMetz's Keygenme #1 by VladMetz

Author:
VladMetz

Language:
C/C++

Upload:
2022-02-01 10:34

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Platform:
Windows

Difficulty:
3.0

Quality:
3.8

Arch:
x86-64

Downloads:
21

Size:
226.50 KB

Writeups:
1

Comments:
4

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 (4)
Writeups (1)

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Sir_Zed on 2022-02-15 17:42: 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 2022-07-04 17:18: 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 2022-07-11 11:27: [Click to reveal]

Marshad on 2022-07-22 11:13: 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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Solution by ThePhilosopher on 2022-07-04 05:55:
solution

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crackmes.one

crackmes.one

VladMetz's Keygenme #1 by VladMetz