-------------------------------lined by aneasystone-----------------------------------------
use the exception of DividedByZero.
at first, the username and sn is requested to be length of 6.
second, we need to know the steps of calculation;
and the process is like following:
(for example: username = "easyst")
and the result = (((('e'^0xc+'a')^0xc+'s')^0xc+'y')^0xc+'s')^0xc+'t' = 0x028d
(for example: sn = "123456")
and then the result of sn is : (((('1'^0x1a+'2')^0x1a+'3')^0x1a+'4')^0x1a+'5')^0x1a+'6' = 0x????
plz look at here:
00401148 . 83F9 06 cmp ecx, 6
0040114B . 7D 23 jge short 00401170
0040114D . 0FBEB40D D8FE>movsx esi, byte ptr [ebp+ecx-128]-------- username
00401155 . 83F0 0C xor eax, 0C
00401158 . 03C6 add eax, esi
0040115A . 8945 DC mov [ebp-24], eax
0040115D . 0FBEB40D D8FD>movsx esi, byte ptr [ebp+ecx-228]-------- sn
00401165 . 83F2 1A xor edx, 1A
00401168 . 03D6 add edx, esi
0040116A . 8955 D8 mov [ebp-28], edx
0040116D . 41 inc ecx
0040116E .^ EB D5 jmp short 00401145
00401170 > 8BC8 mov ecx, eax
00401172 . 2BCA sub ecx, edx--------------------------- ecx=ecx-edx
00401174 . 99 cdq
00401175 . F7F9 idiv ecx-------------------------------- eax=eax/ecx ;HOW ABOUT ECX=0 !
when i began to crack this, i tried so many times,
at the end i found that whatever i input, the result is same. i failed.
and then i realized the importance of the word
EXCEPTION!
i tried to find an exception,so easy one: DividedByZero
and now the question is how to find a sn
whose result after calculting is equal to the result of username
then i take a try to make a keygen like following
but i know it is not the unique one.
the keygen is not so good, shame on myself...
--------------------------------------lined by aneasystone------------------------------------
//keygen.cpp
#include <stdio.h>
#include <stdlib.h>
int main()
{
char username[20] = {0};
char sn[7] = {0};
printf("plz input your name (must length of 6):");
gets(username);
long resultofname = 0;
for(int i = 0; i < 6; i++)
resultofname = (resultofname + username[i]) ^ 0xC;
resultofname ^= 0xC;
long s = resultofname;
for(i = 0; i < 6; i++)
{
if(s > 0x100)
{
s -= 'z';
sn[5-i] = 'z';
s ^= 0x1a;
}
else
{
if(i == 5)
sn[5-i] = s;
else
{
s -= 'A';
sn[5-i] = 'A';
s ^= 0x1a;
}
}
}
printf("the sn is not unique.\nthe sn for you is: %s\n",sn);
return 0;
}
