ctfshow 月饼杯 crypto2_月自圆

def encrypt(m, a, si):
	c=""
	for i in range(len(m)):
		c+=hex(((ord(m[i])) * a + ord(next(si))) % 128)[2:].zfill(2)
	return c
	
if __name__ == "__main__":
	m = '****************************************************flag{*************}'
	assert(len(m)==71)
	a = randint(50,100)
	salt = ''.join(sample(ascii_uppercase, 4))
	si = cycle(salt.lower())
	c=encrypt(m, a, si)
	print(c)

打开文件，从代码内容上来看，密文的形式是明文中每个字节的ascii值*a+未知ascii值%128

a是再50到100之间的数，可以爆破，salt是随机从A~Z之间的4个不重复的字母，可以爆破，关键在于怎么爆破？m的长度是71，测了一下确实是71，说明代码里m中的flag的位置和真正的明文中flag的位置是一样的，位于第53位，由于16进制一位会变成两位，那么对应到密文就是第105位

53和105是从1开始计数，由于代码下表从0开始计数，代码如下：

from itertools import *
from string import *
from random import *
from mitmproxy import sys
c="3472184e657e50561c481f5c1c4e1938163e154431015e13062c1b073d4e3a444f4a5c5c7a071919167b034e1c29566647600c4e1c2956661b6c1f50622f0016317e563546202a"
m = '****************************************************flag{*************}'
#print(m.find('f'))
#52
salt = list(permutations(ascii_uppercase, 4))
for a in range(50,100):
    print('try:',a)
    for j in salt:
        b=''
        t=''
        for k in j:
            b+=k
        b=b.lower()
        for i in range(52,56):
            t+=hex(((ord(m[i])) * a + ord(b[i-52])) % 128)[2:].zfill(2)
        if t == c[104:112]:
            print('a=',a)
            print('si=',j)
            sys.exit()

得到a=67,salt=“JESQ”

那么再用a和salt爆破一下m

from string import *
from itertools import *
a=67
salt = 'jesq'
s = cycle(salt)
c = '3472184e657e50561c481f5c1c4e1938163e154431015e13062c1b073d4e3a444f4a5c5c7a071919167b034e1c29566647600c4e1c2956661b6c1f50622f0016317e563546202a'
flag=''
for i in range(len(c)//2):
    b=c[:2]
    c=c[2:]
    s1=next(s)
    for k in range(32,127):
        c1=hex((k* a + ord(s1)) % 128)[2:].zfill(2)
        if c1==b:
            flag+=chr(k)
            break
print(flag)
#now_is_7fad9fcb-d361-4964-821c-177c906b8d20_flag_is_flag{md5(now-salt)}

now_is_7fad9fcb-d361-4964-821c-177c906b8d20_flag_is_flag{md5(now-salt)}

意思是，now is 7fad9fcb-d361-4964-821c-177c906b8d20

flag是经过md5加密的now-salt，salt是JESQ，那么now-salt就是

7fad9fcb-d361-4964-821c-177c906b8d20-JESQ，

扔到md5里加密一下，得到  1efce62ee0a96e39149e2179db1dd04c