## Description
Warmup your crypto skills with the superior number system!

* enc.py
* flag.enc

## Write up
when looking at `enc.py`
```
fib = [1, 1]
for i in range(2, 11):
fib.append(fib[i - 1] + fib[i - 2])

def c2f(c):
n = ord(c)
b = ''
for i in range(10, -1, -1):
if n >= fib[i]:
n -= fib[i]
b += '1'
else:
b += '0'
return b

flag = open('flag.txt', 'r').read()
enc = ''
for c in flag:
enc += c2f(c) + ' '
with open('flag.enc', 'w') as f:
f.write(enc.strip())

```
you can see that every charachter of the flag is being encrypted through a function called c2f through transforming the ascii code n by comparing it to the fibonacci number corresponding to that index and replacing it by b..u get the idea

## My Approach
My approach is mathematically based and it consists basically on finding the lower and upper bound of n at each iteration of the loop:
by iterating through the binary if the charachter is 0 then n was at the iteration less than that fib[i] so u can at each time change the lower bound and upper bound
so u can finally find lower and upper bound differing by just 1 and since the comparison in the enc.py is "greater or equal" so n at the last iteration is the lower bound
and then I apply the chr function to get the charachter put together u get the flag :D

## Code

```
fib = [1, 1]
for i in range(2, 11):
fib.append(fib[i - 1] + fib[i - 2])

with open('flag.enc','rb') as f:
ct=f.read()
ct_blocks = ct.split(b' ')
pt=''
for blo in ct_blocks:
upper=0
lower=0
test_n=0
for b,i in zip(blo,range(10,-1,-1)):
idx = bytes([b])
if idx == b'1':
lower= fib[i]-test_n
test_n-=fib[i]
else:
upper= fib[i]-test_n
print(lower,upper)
pt+=chr(lower)
print(pt)

#FLAG: corctf{b4s3d_4nd_f1bp!113d}

```

#hmxforlife