A slight reduction, a slight change. from Crypto....

A slight reduction, a slight change.

from Crypto.Util.number import *
import uuid

flag = "moectf{" + str(uuid.uuid4()) + "}"
l = len(flag)
rounds = l - 13
coef = [ord(i) << 248 for i in flag]
p = getPrime(256)
print(f'{p = }')

def eval_poly(x):
y = sum([c * pow(x, i, p) for i, c in enumerate(coef)]) % p
return y

def query():
try:
x = int(input("input x: "))
assert 0 < x < p
except:
print("something wrong")
exit()
print(eval_poly(x))

menu = '''1. query
2. quit'''

for i in range(rounds):
print(f'ROUND {i + 1}/{rounds}')
print(menu)
op = int(input('choice: '))
if op == 1:
query()
if op == 2:
exit()
print("All rounds finished. It's your time to find the flag!")

┌──(kali㉿kali)-[~/crypto/moectf2026/flag/.git]
└─$ nc 192.168.0.5 52611

                         ____            _           ____  

_ __ ___ ___ ___ | _ \ ___ | | _ _ |___ \
| '_ ` _ \ / _ \ / _ \ | |) | / _ \ | | | | | | __) |
| | | | | | | () | | / | / | () | | | | || | / /
|| || |_| _/ _| |_| _/ || _, | |__|
|/

p = 68593641001879686723306593344861817944734016088775804580436696656471711663027
ROUND 1/31

1. query
2. quit
   choice:

帮我完成一下这道ctf挑战,直接输出完整的解题代码并且详细解释