试图用AES解密文件，但会出错：填充不正确

Question

我得到了此代码，该代码将加密字符串“标志”。

from Crypto.Util.number import getPrime
from Crypto.Util.Padding import pad, unpad
from Crypto.Cipher import AES
import hashlib

prime = getPrime(1024)
privkey = random.randint(2, prime - 1)
key = pow(2, privkey, prime)

sk = pow(key, privkey, prime)
aes_key = hashlib.md5(str(sk).encode()).digest()
cipher = AES.new(aes_key, AES.MODE_ECB)
pt = cipher.encrypt(pad(flag, 16)).hex()
        
print(f"[+] This encrypted flag : {pt}")

我被告知要解密加密的字符串以获取原始字符串。我被赋予了加密标志和键的值，以及十六进制中的Prime和Prime键（Privkey）的值。我试图使用这些值这样的标志这样：

flag = "8fceb2a29cc2d7abd8ecfc8da5dc1eea6f67f7a0b047749d66ef8886bb33c720dfc5dd4e508bd1e4a811c62b83f98e65"

prime = int("0xf9aecd571c9afadaceae0004000c64fceb6720f717756dab1f12b2ed7fd211a13024735efeb80a8f7982a0787d4a2eb866b18b8e7d62f2b92f6bd0d7ca52b2cd18e7b508d1af3c69eee907ab9bde2cca7f6cea613954d98a3f8e0c52761937636afb2b6776ac7f4ac02af12e72f4f4905dbeac3e4e856c8542bbda24106161d9", 16)

privkey = int("0x3e1591ea4e4eef19c99626ab1d15d442becbbd2b7d7a4150ee8f1af3f0adf9df47a53823ddfe83c6a7fa4b1b5dfa319021b26dec15c385d3869c7a7ce039b8519318563602d846ea242550bbac73dfc20a27c19b119820e45589cc6f54e9bafc50befbe222aa2738a35f5fca17ca7eec71ce24449ed21fd46b92ca11080001", 16)

key= 101752188851588702786663864886064578902654651951985866839003796634186954471878272123772894282171928731095228234190527287304860559135921159182420718259970442394992811637314757293507073993913485850566751318782466533493182193918336800513466736844109978537994535285068729297204514757610248021028835645897421370304

sk = pow(key, privkey, prime)
aes_key = hashlib.md5(str(sk).encode()).digest()
cipher = AES.new(aes_key, AES.MODE_ECB)
ct = unpad(cipher.decrypt(bytes.fromhex(flag)), 16)

print(f"[+] This decrypted flag : {ct}")

但这似乎不起作用，因为我一直在变得

valueerror：填充不正确。

我在哪一部分错误？

---

编辑： privkey的描述：

Answer 1

问题不是解密代码，而是不完整的密钥privkey。

屏幕截图中的密钥的描述与 privkey泄露以及两个尾随下划线以及这是黑客马拉松的挑战的信息，使我怀疑钥匙是不完整的，需要补充两个十六进制的数字，全部键要确定。

如果将一个字节添加到循环中从0到255的密钥的末端添加一个字节，并且PKCS＃7填充被用作成功解密的标准，则可以确认此假设。如果完成此操作，则结果是明文：

CDDC22{D1ffi3_H3llm4n_k3y_3xch@ng3_D0ne!}

privkey：

3e1591ea4e4eef19c99626ab1d15d442becbbd2b7d7a4150ee8f1af3f0adf9df47a53823ddfe83c6a7fa4b1b5dfa319021b26dec15c385d3869c7a7ce039b8519318563602d846ea242550bbac73dfc20a27c19b119820e45589cc6f54e9bafc50befbe222aa2738a35f5fca17ca7eec71ce24449ed21fd46b92ca11080001d6

IE 0xD6作为最终字节

---

完整代码：

from Crypto.Util.Padding import unpad
from Crypto.Cipher import AES
import hashlib

prime = int("0xf9aecd571c9afadaceae0004000c64fceb6720f717756dab1f12b2ed7fd211a13024735efeb80a8f7982a0787d4a2eb866b18b8e7d62f2b92f6bd0d7ca52b2cd18e7b508d1af3c69eee907ab9bde2cca7f6cea613954d98a3f8e0c52761937636afb2b6776ac7f4ac02af12e72f4f4905dbeac3e4e856c8542bbda24106161d9", 16)
privkey = int("0x3e1591ea4e4eef19c99626ab1d15d442becbbd2b7d7a4150ee8f1af3f0adf9df47a53823ddfe83c6a7fa4b1b5dfa319021b26dec15c385d3869c7a7ce039b8519318563602d846ea242550bbac73dfc20a27c19b119820e45589cc6f54e9bafc50befbe222aa2738a35f5fca17ca7eec71ce24449ed21fd46b92ca1108000100", 16)
key= 101752188851588702786663864886064578902654651951985866839003796634186954471878272123772894282171928731095228234190527287304860559135921159182420718259970442394992811637314757293507073993913485850566751318782466533493182193918336800513466736844109978537994535285068729297204514757610248021028835645897421370304
ct = "8fceb2a29cc2d7abd8ecfc8da5dc1eea6f67f7a0b047749d66ef8886bb33c720dfc5dd4e508bd1e4a811c62b83f98e65"

for val in range(255):

    sk = pow(key, privkey, prime)
    aes_key = hashlib.md5(str(sk).encode()).digest()
    cipher = AES.new(aes_key, AES.MODE_ECB)
    try: 
        ptPadded = cipher.decrypt(bytes.fromhex(ct)) 
        pt = unpad(ptPadded, 16)
        print(f"[+] This decrypted flag (padded)      : {ptPadded}")
        print(f"[+] This decrypted flag (unpadded)    : {pt}")
        print(f"[+] This decrypted flag (UTF8 decoded): {pt.decode('utf8')}")
        print(f"[+] privkey (hex)                     : {(privkey).to_bytes(128, byteorder='big').hex()}")
    except:
        pass
    
    privkey += 1

输出：

[+] This decrypted flag (padded)      : b'CDDC22{D1ffi3_H3llm4n_k3y_3xch@ng3_D0ne!}\n\x06\x06\x06\x06\x06\x06'
[+] This decrypted flag (unpadded)    : b'CDDC22{D1ffi3_H3llm4n_k3y_3xch@ng3_D0ne!}\n'
[+] This decrypted flag (UTF8 decoded): CDDC22{D1ffi3_H3llm4n_k3y_3xch@ng3_D0ne!}

[+] privkey (hex)                     : 3e1591ea4e4eef19c99626ab1d15d442becbbd2b7d7a4150ee8f1af3f0adf9df47a53823ddfe83c6a7fa4b1b5dfa319021b26dec15c385d3869c7a7ce039b8519318563602d846ea242550bbac73dfc20a27c19b119820e45589cc6f54e9bafc50befbe222aa2738a35f5fca17ca7eec71ce24449ed21fd46b92ca11080001d6