Warmup your crypto skills with the superior number system!

We're provided two files -- an encryption script in python and an encrypted flag.

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())

flag.enc

10000100100 10010000010 10010001010 10000100100 10010010010 10001000000 10100000000 10000100010 00101010000 10010010000 00101001010 10000101000 10000010010 00101010000 10010000000 10000101000 10000010010 10001000000 00101000100 10000100010 10010000100 00010101010 00101000100 00101000100 00101001010 10000101000 10100000100 00000100100

Each character is "encrypted" individually which means it's looking awfully tractable. We can engage in a little wholesome brute force to crack each character individually.

from z3 import *
from string import printable
fib = [1, 1]
for i in range(2, 11):
    fib.append(fib[i - 1] + fib[i - 2])
def brute(target):
    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
    for x in printable:
        if c2f(x) == target:
            return x
flag = open('flag.enc', 'r').read().split(" ")
for i in flag:
    char = brute(i)
    print(char,end="")

The flag is corctf{b4s3d_4nd_f1bp!113d}