2023 ImaginaryCTF - snailchecker

Overview

• Points: 100

Description

Optimize me, if you dare. Or not. It might run if you try hard enough.

Attached

snailchecker.py

Solution

By allforest01

def josephus(n):
    return 2 * (n - 2 ** (n.bit_length() - 1)) + 1

def enc1(b):
    n = b[0] * 2 ** 24 + b[1] * 2 ** 16 + b[2] * 2 ** 8 + b[3]
    return josephus(n) - 1

def enc2(n):
    return bytes.fromhex(hex(enc1(n[::-1]))[2:].zfill(8))

charset = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/_{}'
table = dict()
cnt = 0
for i in charset:
    for j in charset:
        for k in charset:
            for l in charset:
                s = (i + j + k + l).encode()
                table[enc2(s)] = s
    cnt += 1
    print(cnt / len(charset) * 100)

flag = b'L\xe8\xc6\xd2f\xde\xd4\xf6j\xd0\xe0\xcad\xe0\xbe\xe6J\xd8\xc4\xde`\xe6\xbe\xda>\xc8\xca\xca^\xde\xde\xc4^\xde\xde\xdez\xe8\xe6\xde'
out = b''
for n in [flag[i:i+4] for i in range(0,len(flag),4)]:
  out += table[n]
print(out)

The flag is

ictf{josephus_problem_speed_boooooooost}