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Version : Python 3.9

import dis
import struct

def MX(y, z, sum, k, p, e):
return (z >> 5 ^ y >> 2) + (y << 3 ^ z << 4) ^ (sum ^ y) + (k[p & 3 ^ e] ^ z)

def btea(v, n, k):

text
u32 = lambda x: x & 0xFFFFFFFFL
y = v[0]
sum = 0
DELTA = 1163219540
if n > 1:
    z = v[n - 1]
    q = 6 + 52 // n
    if q > 0:
        q -= 1
        sum = u32(sum + DELTA)
        e = u32(sum >> 2) & 3
        p = 0
        if p < n - 1:
            y = v[p + 1]
            z = v[p] = u32(v[p] + MX(y, z, sum, k, p, e))
            p += 1
            continue
        y = v[0]
        z = v[n - 1] = u32(v[n - 1] + MX(y, z, sum, k, p, e))
        continue
    return True

if name == 'main':
print('WOW')

Visit https://www.lddgo.net/string/pyc-compile-decompile for more information

Version : Python 3.9

import struct
from crypto import *
from sys import *
import base64
import myalgo
welcome_msg = 'V2VsYzBtMyBUbyBUaGUgV29ybGQgb2YgTDFsYWMgPDM='
input_msg = ':i(G#8T&KiF<F_)F`JToCggs;'
right_msg = 'UmlnaHQsIGNvbmdyYXR1bGF0aW9ucyE='
wrong_msg = 'V3JvbmcgRmxhZyE='
print(b64decode(welcome_msg).decode())
flag = input(a85decode(input_msg).decode())
if not flag.startswith('LilacCTF{') and flag.endswith('}') or len(flag) == 26:
print(b64decode(wrong_msg).decode())
else:
flag = flag[9:25]
res = [
761104570,
1033127419,
0xDE446C05L,
795718415]
key = struct.unpack('<IIII', b'1111222233334444')
input = list(struct.unpack('<IIII', flag.encode()))
myalgo.btea(input, 4, key)
if input[0] == res[0] and input[1] == res[1] and input[2] == res[2] and input[3] == res[3]:
print(b64decode(right_msg).decode())
else:
print(b64decode(wrong_msg).decode())

flag是LilacCTF{}包裹的，给我脚本得到flag