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CrackMe Kinda basic crackme(me)

  • Автор темы Автор темы Rxvan
  • Дата начала Дата начала
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Bad string encryption,lul(2 minutes for patch(rva:6093F))
Screenshot_966.png
 
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you guys only know how to patch a single asm instruction, you are supposed to get the password, thats the point of this crackme, you arent trying to get the end message.... facepalm x10000
 
Обратите внимание, пользователь заблокирован на форуме. Не рекомендуется проводить сделки.
you guys only know how to patch a single asm instruction, you are supposed to get the password, thats the point of this crackme, you arent trying to get the end message.... facepalm x10000

Who cares about your virtualized code , if for patch your crackme need one minute :FailFish: x1000
 
hey dude, get it that the dudes who pissed on your program in less than 5 minutes won't get any feedback from it in terms of xD experience and as a result won't waste their time on it. shove your algorithm up your ass that only a blind person can't make out
so it will be clearer what exactly to shove up your ass
1665399283637.png
 

Вложения

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the point of this isnt to patch it, like i said, get the password or you failed. :joycat:
you didn't even mention that guys need to get password. and in real life no one will try to find it, they will use the simplest way of cracking the software.

what about improving security so ppl won't be able to crack it with 1 patch?
 
the point of this isnt to patch it, like i said, get the password or you failed. :joycat:


C++:
Expand Collapse Copy
std::string pass = password;

int xor_key = -2128831035;
int mul = 16777619;
int result = 0;

for(std::size_t i{0} ; i < pass.size ( ) ; ++i)
{
    result = (password[i] ^ xor_key) * mul;
    
    xor_key = result;
}

if(result == -626372842)
    gg;
else
    fuck u;

steel allow yourself a lot for algo in 2 lines
 
C++:
Expand Collapse Copy
std::string pass = password;

int xor_key = -2128831035;
int mul = 16777619;
int result = 0;

for(std::size_t i{0} ; i < pass.size ( ) ; ++i)
{
    result = (password[i] ^ xor_key) * mul;
  
    xor_key = result;
}

if(result == -626372842)
    gg;
else
    fuck u;

steel allow yourself a lot for algo in 2 lines
fnv hash and he wants us to find password XDD what a joke
 
the point of this isnt to patch it, like i said, get the password or you failed. :joycat:
This is getting kind of a funny discussion, so I'll say an important point:
The reason why no one wants to look up your password/recover algorithm is the use of CV and literally no originality (not an interesting anti-debug / anti-reverse trick). What prevents me from making a crackme in 5 minutes and covering it with a licensed version of VMP and CV , writing the complexity 1337/10? :kappa:
It would be possible to make it impossible to output the win information if you just used the entered password to decrypt the shellcode (the simplest messagebox) of the win...
It's just not interesting :LUL: ...
 
you are all retarded, there is no checks, why are you flexing that you can patch a function, its literally 1 asm instruction.... you are completely missing the point of this, COMPLETELY, stop it bruh.
Посмотреть вложение 224181

Your fucking result not completely divided man , thats enough to understand what u are idiot
?????????
C++:
Expand Collapse Copy
std::string pass = password;

int xor_key = -2128831035;
int mul = 16777619;
int result = 0;

for(std::size_t i{0} ; i < pass.size ( ) ; ++i)
{
    result = (password[i] ^ xor_key) * mul;
   
    xor_key = result;
}

if(result == -626372842)
    gg;
else
    fuck u;

steel allow yourself a lot for algo in 2 lines
read what i said before, it isnt a string, there is no xoring.... why are you xoring it?
you didn't even mention that guys need to get password. and in real life no one will try to find it, they will use the simplest way of cracking the software.

what about improving security so ppl won't be able to crack it with 1 patch?
i wrote it in 2 minutes, im sorry you are all so smooth brained, i literally said 4 times you need to find the password, idk why you all think patching is the only form of cracking... what happens when they update it and fix the patches but the password is still the same? then what are you gonna do...
 
you are all retarded, there is no checks, why are you flexing that you can patch a function, its literally 1 asm instruction.... you are completely missing the point of this, COMPLETELY, stop it bruh.

?????????

read what i said before, it isnt a string, there is no xoring.... why are you xoring it?

i wrote it in 2 minutes, im sorry you are all so smooth brained, i literally said 4 times you need to find the password, idk why you all think patching is the only form of cracking... what happens when they update it and fix the patches but the password is still the same? then what are you gonna do...

1665503602806.png


there is no xoring XDD ))

cannot get pass if it was hashed man , try to understand this.......

I show u what needed result dont divided completely
1665503705428.png


imul esi , ecx , 0x1000193 == int esi = ecx * 0x1000193;
if esi == 0xdaaa4f16
good ;


Approval : for "Good pass" u need last number in ESI mul with 0x1000193 and get 0xDAAA4F16 , u will never get this number man , NEVER , try to understand
 
you are all retarded, there is no checks, why are you flexing that you can patch a function, its literally 1 asm instruction.... you are completely missing the point of this, COMPLETELY, stop it bruh.

?????????

read what i said before, it isnt a string, there is no xoring.... why are you xoring it?

i wrote it in 2 minutes, im sorry you are all so smooth brained, i literally said 4 times you need to find the password, idk why you all think patching is the only form of cracking... what happens when they update it and fix the patches but the password is still the same? then what are you gonna do...
a sick piece of shit on your head, they answered you yesterday that they were not going to look for a password from your shit, it's not interesting to anyone, admit that you fucked up even more when you were thrown off the disassembled algorithm and the result of division, which will never be whole
 
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