Core Decryption was another one of those painful problems. Though not hard, it's hard to find what direction you need to take. First off, you can't just run GDB with the core file. You have to compile the cryptor.c with gcc -m32 -g cryptor.c -o cryptor, we compile it in 32 bit mode with debugging symbols ON. This is the key to getting the core dump running. Running the core file and binary through GDB with the following commands:

root@neuron:~$ gdb cryptor core.26474
(gdb) bt
#0  0x08048564 in swap (sbox=0xffffd17c "", i=-11652, j=16) at cryptor.c:7
#1  0x08048719 in crypt (inf=0x0, keyf=0xffffd17c, outf=0xf7ffda5c) at cryptor.c:42
#2  0x08048894 in main (argc=4, argv=0xffffd374) at cryptor.c:67
(gdb) f 1
#1  0x08048719 in crypt (inf=0x0, keyf=0xffffd17c, outf=0xf7ffda5c) at cryptor.c:42
42 tmp ^= next_prg(sbox,&i,&j);
(gdb) info locals
sbox = "\000\001\002\003\004\005\006\a\b\t\n\v\f\r\016\017\020\021\022\023\024\025\026\027\030\031\032\033\034\035\036\037 !\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~\177\200\201\202\203\204\205\206\207\210\211\212\213\214\215\216\217\220\221\222\223\224\225\226\227\230\231\232\233\234\235\236\237\240\241\242\243\244\245\246\247\250\251\252\253\254\255\256\257\260\261\262\263\264\265\266\267\270\271\272\273\274\275\276\277\300\301\302\303\304\305\306\307\310\311\312\313\314\315\316\317\320\321\322\323\324\325\326\327\330\331\332\333\334\335\336\337\340\341\342\343\344\345\346\347\350\351\352\353\354\355\356\357\360\361\362\363\364\365\366\367\370\371\372\373\374\375\376\377"
i = 0
j = -135168078
key = "\305\344L\004\273/\\\020\272u\033×L\334", 
tmp = 247 '\367'

So we see that there is a key variable here. At first, we wondered what to do with it, what did "incomplete sequence" mean? We decided to ignore it, because if you refer back to the code, you know that the key will be 16 bytes long. We took the key, turned into hex so Python wouldn't have any trouble with it, and we got our result.

import random, base64
def crypt(data, key):
    x = 0
    box = range(256)
    for i in range(256):
        x = (x + box[i] + ord(key[i % len(key)])) % 256
        box[i], box[x] = box[x], box[i]
    x = y = 0
    out = []
    for char in data:
        x = (x + 1) % 256
        y = (y + box[x]) % 256
        box[x], box[y] = box[y], box[x]
        out.append(chr(ord(char) ^ box[(box[x] + box[y]) % 256]))
    return ''.join(out)
with open('ENCRYPTED', 'rb') as f:
        content =
print crypt(content, 'c5e44c04bb2f5c10ba751bc3974cdcdc'.decode('hex'))

The message, decrypted with RC4, was:

Well, assuming you're able to actually keep the secret keyfile safe, this is a legitimate way to encrypt files!
Your key is: astronomy_ceremony_times



温馨提示: 此处内容需要评论本文后刷新才能查看,支付2元即可直接查看所有Flag。