3-Cipher#

題目

3-cipher.py는 Key1, Key2, AES_iv, AES_cipher_text를 출력하는 암호화 프로그램입니다.
AES_cipher_text는 플래그를 암호화한 문자열이고, Key1, Key2, AES_iv는 플래그 복호화에 필요한 값입니다.
AES_cipher_text를 복호화하여 플래그를 구하세요. 플래그는 FLAG 변수에 있습니다.
플래그 형식은 DH{…} 입니다.

3-cipher.py

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#!/usr/bin/env python3
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from Crypto.Cipher import AES
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from Crypto.Random import get_random_bytes
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from base64 import b64encode
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from Crypto.Util.Padding import pad
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import string
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try:
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    FLAG = open("flag", "rb").read()        # flag is here!
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except:
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    FLAG = b'[**FLAG**]'
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class AES_Cipher:
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    def __init__(self):
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        self.key = get_random_bytes(16)
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    def encrypt(self, data: bytes):
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        cipher = AES.new(self.key, AES.MODE_CBC)
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        self.iv = b64encode(cipher.iv).decode('utf-8')
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        self.cipher_t = b64encode(cipher.encrypt(pad(data, AES.block_size))).decode('utf-8')
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        return 'AES_iv: ' + self.iv + '\nAES_cipher_text: ' + self.cipher_t
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class RSA_Cipher:
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    def __init__(self):
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        self.e = 65537
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        self.n = 13119132709177697801
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    def encrypt(self, data: int):
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        cipher_n = pow(data, self.e, self.n)
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        return cipher_n
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class Caesar_Cipher:
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    def __init__(self):
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        self.alpha = list(string.ascii_lowercase)
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    def encrypt(self, data: str):
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        cipher_t = [i for i in range(len(data))]
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        for i in range(len(data)):
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            ch = data[i]
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            if(ch in self.alpha):
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                idx = self.alpha.index(ch)
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                cipher_t[i] = self.alpha[(idx + 13) % 26]
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            else:
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                cipher_t[i] = ch
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        return ''.join(cipher_t)
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ac = AES_Cipher()
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rc = RSA_Cipher()
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cc = Caesar_Cipher()
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ac_key = int.from_bytes(ac.key, 'big')
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# Key 1
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rc_p = hex(ac_key)[2:16]
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rc_c = rc.encrypt(int(rc_p, 16))
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print('Key1:', hex(rc_c)[2:])
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# Key 2
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cc_p = hex(ac_key)[16:]
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cc_c = cc.encrypt(cc_p)
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print('Key2:', cc_c)
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# Result
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print(ac.encrypt(FLAG))

輸出

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Key1: a2261df72c92c922
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Key2: 26278o6415949oqn06
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AES_iv: 0FVXK8rOgrGWpXSrTAhDYg==
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AES_cipher_text: qqD5QHpddaV0QxWsfeTa3A==

Key1 是 ac_key 的前 14 個字符，用 RSA 加密過

Key2 是 ac_key 後半部分，用 ROT13 (Caesar Cipher) 加密過

Key1 + Key2 會獲得完整 ac_key

用 AES.new(ac.key, AES.MODE_CBC, iv) 進行 AES 解密

用 unpad 移除填充，取得 FLAG

把 n 拿去 factordb

factorDB

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p = 3573716833
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q = 3671005097

script

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from base64 import b64decode
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from Crypto.Cipher import AES
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from Crypto.Util.Padding import unpad
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from Crypto.Util.number import long_to_bytes
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import string
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KEY1 = "a2261df72c92c922"
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KEY2 = "26278o6415949oqn06"
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AES_IV = b64decode("0FVXK8rOgrGWpXSrTAhDYg==")
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AES_CIPHER_TEXT = b64decode("qqD5QHpddaV0QxWsfeTa3A==")
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e = 65537
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n = 13119132709177697801
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p = 3573716833
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q = 3671005097
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def decrypt_rsa(cipher_text_hex):
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    cipher_text = int(cipher_text_hex, 16)
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    d = pow(e, -1, (p - 1) * (q - 1))
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    decrypted_int = pow(cipher_text, d, n)
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    return long_to_bytes(decrypted_int)
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def rot13_decrypt(cipher_text):
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    alpha = string.ascii_lowercase
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    decrypted = []
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    for ch in cipher_text:
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        if ch in alpha:
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            idx = alpha.index(ch)
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            decrypted.append(alpha[(idx - 13) % 26])
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        else:
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            decrypted.append(ch)
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    return ''.join(decrypted)
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Key1_decrypted = decrypt_rsa(KEY1).hex()
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Key2_decrypted = rot13_decrypt(KEY2)
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aes_key_hex = Key1_decrypted + Key2_decrypted
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aes_key = bytes.fromhex(aes_key_hex)
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cipher = AES.new(aes_key, AES.MODE_CBC, iv=AES_IV)
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flag = unpad(cipher.decrypt(AES_CIPHER_TEXT), AES.block_size)
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print("FLAG:", flag.decode())

執行結果

test result

開虛擬機後得到的題目是

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Key1: 5dcf8df268608c3
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Key2: s380r0p091qs7q9sos
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AES_iv: eSnk/TTlXp4ZgHxIKEXmYw==
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AES_cipher_text: q5olGh0B/7TloXWgZqtfvjF85AY1wA08Seq/+arA2DDzF7BhFRMrlpowvZJcfNBfR9oj3bnd6nqMwy2woe56EwlNpBNkcdQEzhfn+RCPWp8=

執行 script 結果

flag

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DH{baeeb7554473c36d05ae660192e685665c48cc8edcb6e209e15aa44cc957dc20}