Decryption problem

[xulescu]

Hi! I'm working on something and hit a dead end. I have to try and find the algorythm that decrypts some DWORDs into readable strings. Any help would be greatly appreciated. Here's what I have so far:
I do know that the following hex seqences mean their respective strings:

Code: Select all

F6   58   C8   73   00   00   00   00   berlin
C9   F2   5C   41   22   5C   00   00   amsterdam
4C   79   19   37   00   00   00   00   lisboa
CF   6F   14   00   00   00   00   00   wien
CB   54   98   63   3C   01   00   00   warszawa
F8   80   CA   B6   0F   00   00   00   brussel
76   B8   14   00   00   00   00   00   bern
2E   50   6D   01   00   00   00   00   praha
56   BB   09   00   00   00   00   00   roma
A9   1F   17   79   00   00   00   00   milano
C1   BB   92   44   00   00   00   00   madrid
B6   BD   EC   93   10   2E   00   00   barcelona
52   AB   1B   AF   9F   03   00   00   bordeaux
8C   AA   14   00   00   00   00   00   lyon
04   37   AB   03   00   00   00   00   paris
49   75   4D   1C   11   00   00   00   munchen
94   28   7A   EB   A0   79   00   00   frankfurt
53   43   6C   04   00   00   00   00   italy
29   F0   6F   13   08   00   00   00   austria


and the list can continue...
If you notice any patterns or procedures or can just give me a hint if not a solution I would still grately appreciate. Thanks!

[r-ID]

#include <stdlib.h>

char buffer[32];
ltoa((long)dwValue, buffer, 10);

[xulescu]

Thanks for trying, but it doesn't work. I've tried ltoa in all bases, same result random combination of letters and numbers, nowhere near good result.

[r-ID]

are you able to enter your own strings? for example, a string like aaaa, whould be really helpfull

-- Tue Jul 14, 2009 4:34 am --

i don't see any pattern. if it uses any encryption, it is almost imposible to reverse an algo. And hex doesn't look like simple encoding to me.

[xulescu]

I am able to give it any hex input and find out the decrypted string.

Also, something else I know that might be helpfull:

Code: Select all

0x03CF0508 = "paint"
0x03CF0508 + 0x4B90860 * 1 = "paint1"
0x03CF0508 + 0x4B90860 * 2 = "paint2"
.....
0x03CF0508 + 0x4B90860 * 10 = "paint0"
0x03CF0508 + 0x4B90860 * 11 = "painta"
0x03CF0508 + 0x4B90860 * 12 = "paintb"
......