Can-CWIC CTF 2017 rev me easy writeup

hi there 

in this article i'll tell you how i solved the rev me easy 

challenge in can-cwic ctf 2017,so let's go

first the challenge look like the image below

 

after downloading the file 

Let’s take a look at the binary in IDA

  

so as we saw it is pretty simple 

it is just move group of values one by one to edx then xor it with the value in eax 

now what is the value in eax 

let's scroll up a little bit and we found it it is 0x10

so after i collect all the values that been mov to edx and xored 

in order to solve the challenge i wrote a python script that do 

the same operation the app do and print the result  wish supposed to

be the flag

cc = ""
bb = []
aa = [0x56,0x5c,0x51,0x57,0x6b,0x42,0x23,0x66,0x75,0x62,0x63,0x79,0x7e,0x77,0x4f, 0x79,0x25,0x4f,0x7e,0x20,0x64,0x4f,0x64,0x78,0x71,0x64,0x4f,0x78,0x71,0x62,0x74,0x2f,0x6d,0x7]
for a in aa:
    bb.append(a^0x10)

for b in bb:
    cc += (chr(b))
print(cc)




first it xor every value in aa with 0x10 the it append the it to bb
wish become the xored values list 
second it will convert every value in bb from hex to char and store
it into cc
then finally it will print cc  
and here is our flag