easyaes

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WriteUp来源

https://xz.aliyun.com/t/8581

题目考点

• AES加密

解题思路

hint和key长度不等,且hint为四个字符的重复

即输出的密文中有部分hint的字符,可以首先恢复hint再逐步次求出key

已知密文、明文、私钥可以根据aes的cbc模式原理求出位移

#!/usr/bin/python
from Crypto.Util.number import long_to_bytes
import binascii, sys
from Crypto.Util.strxor import strxor
from Crypto.Cipher import AES

# -----------get key---------
tmp = 56631233292325412205528754798133970783633216936302049893130220461139160682777
hint = int(str(hex(tmp))[2:10] * 8,16)
key = long_to_bytes(tmp ^ hint)

# ----------get iv-----------
msg = b'Welcome to this competition, I hope you can have fun today!!!!!!'
msgs = [msg[ii:(ii+16)] for ii in range(0,len(msg),16)]
msgs.reverse()
IV = binascii.unhexlify('3c976c92aff4095a23e885b195077b66')

def decry(key,IV,ms):
    aes=AES.new(key,AES.MODE_ECB)
    return strxor(aes.decrypt(IV),ms)

for ms in msgs:
    IV=decry(key,IV,ms)
print(b'd0g3{' + IV+ b'}')

Flag

d0g3{aEs_1s_SO0o_e4sY}