Author:
VladMetz

Language:
C/C++

Upload:
10:05 PM 02/19/2022

Platform
Windows

Difficulty:
4.0

Quality:
3.5

Arch:
x86

Description

I think the difficulty here is not in the anti-debugging methods Good luck :D

• Comments
• Solutions

nogame on 9:46 AM 02/23/2022: if((key[14] == key[13]) && (key[10] - key[9] = 0)) { if(key[12] - key[11] = 0) { if(4 * (key[10] - key[9]) * (key[12] - key[11]) * (key[8] - key[7]) == -2912) { // Nice, good work :) } } }

Shert on 2:48 PM 03/02/2022: Спасибо. Хороший семпл :-) Кей-ген отправил... Your key = vladxox100J4P99 Your key = vlaexox100J4P99 Your key = vladxox100J4P:: Your key = vlaexox100J4P:: Your key = vladxox100J4P;; Your key = vlaexox100J4P;; Your key = vladxox100J4P vlaexox100J4P vladxox100J4P== Your key = vlaexox100J4P== Your key = vladxox100J4P Your key = vlaexox100J4P Your key = vladxox100J4P?? Your key = vlaexox100J4P?? Your key = vladxox100J4P@@ Your key = vlaexox100J4P@@ Your key = vladxox100J4PAA Your key = vlaexox100J4PAA Your key = vladxox100J4PBB Your key = vlaexox100J4PBB

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