安洵杯 2020easyaes

#!/usr/bin/python
from Crypto.Cipher import AES
import binascii
from Crypto.Util.number import bytes_to_long,long_to_bytes
from flag import flag
from key import key

iv = flag.strip(b'd0g3{').strip(b'}')

LENGTH = len(key)
assert LENGTH == 16

hint = os.urandom(4) * 8
print(bytes_to_long(hint)^bytes_to_long(key))

msg = b'Welcome to this competition, I hope you can have fun today!!!!!!'

def encrypto(message):
    aes = AES.new(key,AES.MODE_CBC,iv)
    return aes.encrypt(message)

print(binascii.hexlify(encrypto(msg))[-32:])

'''
56631233292325412205528754798133970783633216936302049893130220461139160682777
b'3c976c92aff4095a23e885b195077b66'
'''

这道题涉及AES的ECB加密

因为hint是4字节重复8次，共32字节，而key只有16字节，异或后只有后16字节改变了可以通过前16字节获取hint，再异或回去就可以获得key了

hint='{:<4X}'.format(56631233292325412205528754798133970783633216936302049893130220461139160682777)[:4*2]*8
key=long_to_bytes(a^int(hint,16))

根据这题，我们需要获取iv，根据CBC模式和ECB模式的转化可以获取

m=b'xxxxxxxxxxxxxxxxyyyyyyyyyyyyyyyy'
key=b'aEs_1s_SO0o_e4sY'
iv=b'D0g3D0g3D0g3D0g3'
aes=AES.new(key,AES.MODE_CBC,iv)
enc=aes.encrypt(m)
print(enc)

其等效于ECB模式的，第一步第一组明文异或初始向量，第二部第二组明文异或上一次的加密结果，以此类推

from Crypto.Cipher import AES
from Crypto.Util.strxor import strxor
m1=b'xxxxxxxxxxxxxxxx'
m2=b'yyyyyyyyyyyyyyyy'
aes=AES.new(key,AES.MODE_ECB)
enc1=aes.encrypt(strxor(m1,iv))
enc2=aes.encrypt(strxor(m2,enc1))
enc=enc1+enc2
print(enc)

此处msg有64字节，所以有4组，等效于ECB模式的

from Crypto.Cipher import AES
from Crypto.Util.strxor import strxor
msg = b'Welcome to this competition, I hope you can have fun today!!!!!!'
m1=msg[:16]
m2=msg[16:32]
m3=msg[32:48]
m4=msg[48:]
aes=AES.new(key,AES.MODE_ECB)
enc1=aes.encrypt(strxor(m1,iv))
enc2=aes.encrypt(strxor(m2,enc1))
enc3=aes.encrypt(strxor(m3,enc2))
enc4=aes.encrypt(strxor(m4,enc3))
enc=enc1+enc2+enc3+enc4

那我们可以直接从enc4递推回iv

from Crypto.Cipher import AES
from Crypto.Util.strxor import strxor
msg = b'Welcome to this competition, I hope you can have fun today!!!!!!'
m1=msg[:16]
m2=msg[16:32]
m3=msg[32:48]
m4=msg[48:]
enc4=b'3c976c92aff4095a23e885b195077b66'
enc4=long_to_bytes(int(enc4,16))#从16进制字符串转化回原字符串
aes=AES.new(key,AES.MODE_ECB)
enc3=strxor(aes.decrypt(enc4),m4)
enc2=strxor(aes.decrypt(enc3),m3)
enc1=strxor(aes.decrypt(enc2),m2)
iv=strxor(aes.decrypt(enc1),m1)

exp:

from Crypto.Cipher import AES
from Crypto.Util.number import bytes_to_long,long_to_bytes
msg = b'Welcome to this competition, I hope you can have fun today!!!!!!'
a=56631233292325412205528754798133970783633216936302049893130220461139160682777
enc4=b'3c976c92aff4095a23e885b195077b66'
enc4=long_to_bytes(int(enc4,16))
msgs=[bytes_to_long(msg[i:i+16]) for i in range(0,len(msg),16)]
hint='{:<4X}'.format(a)[:4*2]*8
key=long_to_bytes(a^int(hint,16))
aes=AES.new(key,AES.MODE_ECB)
enc=[0]*5
enc[4]=enc4
for i in range(3,-1,-1):
	enc[i]=long_to_bytes(bytes_to_long(aes.decrypt(enc[i+1]))^msgs[i])
print(enc[0])