[BSidesSF2020]decrypto-2

題目
Kerckhoffs’s principle states that “A cryptosystem should be secure even if everything about the system, except the key, is public knowledge.” So here’s our really unbreakable cipher.
flag.svg.enc

import hashlib import struct import sysclass Crypto:def __init__(self, key):if not isinstance(key, bytes):raise TypeError('key must be of type bytes!')self.key = keyself._buf = bytes()self._out = open("/dev/stdout", "wb")def _extend_buf(self):self._buf += self.keydef get_bytes(self, nbytes):while len(self._buf) < nbytes:self._extend_buf()ret, self._buf = self._buf[:nbytes], self._buf[nbytes:]return retdef encrypt(self, buf):if not isinstance(buf, bytes):raise TypeError('buf must be of type bytes!')stream = self.get_bytes(len(buf))return bytes(a ^ b for a, b in zip(buf, stream))def set_outfile(self, fname):self._out = open(fname, "wb")def encrypt_file(self, fname):buf = open(fname, "rb").read()self._out.write(self.encrypt(buf))class HashCrypto(Crypto):def __init__(self, *args, **kwargs):super().__init__(*args, **kwargs)self._blk = self.keyself._blkid = 0def _extend_buf(self):self._blk = hashlib.sha256(self._blk + struct.pack('<I', self._blkid)).digest()self._blkid += 1self._buf += self._blkdef main(argv):if len(argv) not in (3, 4):print("%s <key> <infile> [outfile]" % sys.argv[0])returnargv.pop(0)key = argv.pop(0)inf = argv.pop(0)crypter = HashCrypto(key.encode("utf-8"))if sys.argv:crypter.set_outfile(argv.pop(0))crypter.encrypt_file(inf)if __name__ == '__main__':main(sys.argv)

解題

科克霍夫斯原理聲明“密碼系統應該是安全的，即使系統的所有內容，除了密鑰，都是公共知識。”所以這里是我們的真正牢不可破的密碼。

跟 [BSidesSF2020]decrypto-1差不多

解題代碼是搜到了，但是沒看太懂😕

s=b'<?xml version="1.0" encoding="UTF-8" standalone="no"?>' f=open('flag.svg.enc','rb').read() import hashlib import struct sha0=bytes() x1=bytes(a^b for a,b in zip(s[:32],f[:32])) sha0+=x1 temp=1 for i in range(200):x1=hashlib.sha256(x1+struct.pack('<I',temp)).digest()sha0+=x1temp+=1 flag=bytes(a^b for a,b in zip(sha0,f)) print(flag)#CTF{but_even_I_couldnt_break_IT}

struct.pack用于將Python的值根據格式符，轉換為字符串（因為Python中沒有字節(Byte)類型，可以把這里的字符串理解為字節流，或字節數組）。其函數原型為：struct.pack(fmt, v1, v2, …)，參數fmt是格式字符串，關于格式字符串的相關信息在下面有所介紹。v1, v2, …表示要轉換的python值。

答案

flag{but_even_I_couldnt_break_IT}

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