BUUCTF_Re_[2019红帽杯]easyRE

64位,elf文件

 

 

 

 

查找字符串,发现一大串,看着下面的ABCDEFG就是base64

然后多层解密之后就得到一串网址:

https://bbs.pediy.com/thread-254172.htm

当然,没啥卵用

然后发现continue前面有个加密函数

exp:

s1='Iodl>Qnb(ocy'
num=127
s2='y.i'
s3='d`3w}wek9{iy=~yL@EC'
s=[]
flag=''
for i in range(len(s1)):
    s.append(ord(s1[i]))
s.append(num)
for i in range(len(s2)):
    s.append(ord(s2[i]))
s.append(num)
for i in range(len(s3)):
    s.append(ord(s3[i]))
for i in range(len(s)):
    flag+=chr(s[i]^i)
print(flag)

得到:

Info:The first four chars are `flag`

废话

 

 

 然后这里有些没见过的字符串

进函数看看

 

 

 第十行的if判断

v1和.....&YA1A0进行异或==f

v4=v1

v1和.....A3进行异或==g

其实就是字符的前四位异或之后等于flag

exp

s=['40','35','20','56']
s1='flag'
v4=''
for i in range(4):
    v4+=chr(int(s[i],16)^ord(s1[i]))
print(v4)

得到v4=v1=&YA1

 

 接着这个循环

exp

v4='&YA1'
s=['40','35','20','56','5D','18','22','45','17','2F','24','6E','62','3C','27','54','48','6C','24','6E','72','3C','32','45','5B']
flag=''
for i in range(len(s)):
    flag+=chr(int(s[i],16)^ord(v4[i%4]))
print(flag)

get新知识

((_BYTE *)&v4 + j % 4))==v4[j%4]

flag:

flag{Act1ve_Defen5e_Test}