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
```