rrenamed on 7:57 PM 05/09/2022: Pass: 5up3rH1dd3nPa55w0rd
This probably is a bit lower than 3.0
SMUJ on 8:41 PM 05/30/2022: How can i crack the software when I try to drag it into the dissembler it says i need a password for the zip?
Blepperton on 11:20 AM 06/01/2022: unsafe code. program crashes if input is too long.
Nycz on 3:35 PM 06/03/2022: hint: memory is not written to in correct order
db123 on 3:09 AM 06/20/2022: password:5up3rH1dd3nPa55w0rd
codes:
// length must be 19
// ignore memory leak
char* encrypt_simple(char* text)
{
int length; // r8d
int byte; // ecx
assert(strlen(text) == 19);
int* result = new int[19];
memset(result, 0, 19*sizeof(int));
length = 0;
do
{
byte = text[length];
result[length] = byte % 2 == 0 ? byte + 1 : byte - 1;
++length;
} while (length
1337ReverseEngineer on 11:38 PM 07/19/2022: Very nice crackme. Enjoyed it alot.
Marshad on 1:33 AM 07/22/2022: 5up3rH1dd3nPa55w0rd
Nice challenge
Riria on 7:13 AM 08/02/2022: Learn a lot from it!
f1ref1st on 5:33 PM 11/19/2022: to get the password with the harden password:
lst=[0x33, 0x73,0x70,0x31,0x72,0x48,0x2f,0x64,0x64,0x31,0x6e,0x50,0x5f,0x33,0x33,0x75,0x30,0x72,0x64] # harden password
lst2=[]
for i in lst:
lst2.append(int(i+0x1))
for x in lst2:
if i%2 == 0:
print(chr(i+1), end="")
else:
print(chr(i-1), end="")
#now it will print the password, the algorithm dependence on if the ascii value of the character is even, then add one to it, and if it's odd, then minus one.
CyGhost on 4:23 PM 01/21/2023: https://imgur.com/a/z5FO7Kg i think you need to patch this to find password because it will firstly run code that just display's that password is incorrect and than later it will do a while loop for a correct password