If you analyze the binary, you can understand It's just xor.Following this, you can find flag easily.
```#SimpleCryptoWare Decryptor by sqrtrev
f = open("secretInfo","r")c = f.read()f.close()
for i in range(0,95):	tmp = ''	for x in c:		tmp += chr(ord(x)^i)	if 'ctf' in tmp:		print tmp```