Since you konw something

题目代码如下

from pwn import xor
# The Python pwntools library has a convenient xor() function that can XOR together data of different types and lengths
from Crypto.Util.number import bytes_to_long

key = ?? # extremely short
FLAG = 'flag{????????}'
c = bytes_to_long(xor(FLAG,key))

print("c={}".format(c))

'''
c=218950457292639210021937048771508243745941011391746420225459726647571
'''

又是简单的签到题，出现了上周的老朋友 xor()

此题注释非常重要，希望新生能够充分地利用泄露的信息，即 flag 的前一部分

这道题可以关注的地方是 flag 的格式是明确的 flag{ 开头，结合注释，key 极短，直接把 c 和 flag{ 异或一下看看

from pwn import xor
from Crypto.Util.number import long_to_bytes

flag_head = 'flag{'
c=218950457292639210021937048771508243745941011391746420225459726647571
guess_key = xor(long_to_bytes(c), flag_head)
print(guess_key)
# b"nsnsnL2gVcf/xKa}1MQ8z@m'aa1`t"

可以看到，key 的前一部分是重复的 ns，不难猜测 key 就是 ns

from pwn import xor
from Crypto.Util.number import long_to_bytes
c=218950457292639210021937048771508243745941011391746420225459726647571
key='ns'
flag = xor(long_to_bytes(c),key)
print(flag)
# b'flag{Y0u_kn0w_th3_X0r_b3tt3r}'

希望大家都能签到成功喵 ~