| random |
q is determined by:
q = (rand() + 22) % 26 + 65
We have the following conditional:
q ^ correctBuf[i] | bWrong
If the left hand side of the following expression is ever true, bWrong will be set to 1, which makes the check always return true. You fail one character, you fail all of them.
For the left hand side of the expression to not be true, q must be equals to correctBuf[i] as they're XOR'ed together. The first character in correctBuf is "C", which is 67 in ordinal. Thus the first time around, q must be equals 67. Next time q must be equals "O", which is 79 in ordinal.
The solution involves writing a seed cracker, which then solves for the sequence 69, 79... (aka the "CORRECT" string).
There could be multiple seeds, however the one I found is:
2147139625 |
2024-08-27 20:35 |