中には count.py と output.txtがある
import multiprocessing from Crypto.Cipher import AES from secret import key, flag counter = 0 aes = AES.new(key, AES.MODE_ECB) def chunk(input_data, size): return [input_data[i:i + size] for i in range(0, len(input_data), size)] def xor(*t): from functools import reduce from operator import xor return [reduce(xor, x, 0) for x in zip(*t)] def xor_string(t1, t2): t1 = map(ord, t1) t2 = map(ord, t2) return "".join(map(chr, xor(t1, t2))) def pad(data): pad_byte = 16 - len(data) % 16 return data + (chr(pad_byte) * pad_byte) def worker_function(block): global counter key_stream = aes.encrypt(pad(str(counter))) result = xor_string(block, key_stream) counter += 1 return result def distribute_work(worker, data_list, processes=8): pool = multiprocessing.Pool(processes=processes) result = pool.map(worker, data_list) pool.close() return result def encrypt_parallel(plaintext, workers_number): chunks = chunk(pad(plaintext), 16) results = distribute_work(worker_function, chunks, workers_number) return "".join(results) def main(): plaintext = """The Song of the Count You know that I am called the Count Because I really love to count I could sit and count all day Sometimes I get carried away I count slowly, slowly, slowly getting faster Once I've started counting it's really hard to stop Faster, faster. It is so exciting! I could count forever, count until I drop 1! 2! 3! 4! 1-2-3-4, 1-2-3-4, 1-2, i love couning whatever the ammount haha! 1-2-3-4, heyyayayay heyayayay that's the sound of the count I count the spiders on the wall... I count the cobwebs in the hall... I count the candles on the shelf... When I'm alone, I count myself! I count slowly, slowly, slowly getting faster Once I've started counting it's really hard to stop Faster, faster. It is so exciting! I could count forever, count until I drop 1! 2! 3! 4! 1-2-3-4, 1-2-3-4, 1, 2 I love counting whatever the ammount! 1-2-3-4 heyayayay heayayay 1-2-3-4 That's the song of the Count! """ + flag encrypted = encrypt_parallel(plaintext, 32) print(encrypted.encode("hex")) if __name__ == '__main__': multiprocessing.freeze_support() main()
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
AESのCTRモードで平文＋フラグを暗号化しているだけにみえる。threadで並列処理をしてるのでそのときにcounterの値がうまく同期できていない、みたいなのだと嬉しい。もしflagの暗号化と別の平文のブロックの暗号化で同じ暗号化鍵が使われていたらこれで解ける。
とりあえず既知の平文ブロックを暗号化している鍵を全部取り出して試してみる。
すると次のような結果になった
...
'p4{at_the_end_of'
'_the_day_you_can'
'§\x8dæÞwñ$øX\x07\x81«¤¡Ë\x99'
'\x97\x97úÁkÂ!ê%aýüÃÇ\x9cè'
'\x88ÖóÓzñ3ÿH6\x90\x9a¯\x90û\x86'
'§\x96à×QÊ&îr\x10\x9a\x81\x94¬õ\x8e'
'_only_count_on_y'
'ourself}\x08\x08\x08\x08\x08\x08\x08\x08'
'\x0e0õK\x1e¯\x9aö\x889Ô\x83¡\x16J\x1e'
'!pæO5\x94\x8fç²\x1fÞ\x98\x9a*D\x16'
'Ù\x89hô\x1d\x01Êfµa0Faèîá'
'é\x93të\x012ÏtÈ\x07L\x11\x06\x8e¹\x90'
ということでフラグは p4{at_the_end_of_the_day_you_can_only_count_on_yourself}
