Boston Key Party CTF 2014 Crypto 100

Mind your P's and Q's!
Crypto : 100

The flag has been split into several files, and encrypted under RSA-OAEP. Can you break ALL of the ciphertexts, and reassemble the key?
 http://bostonkeyparty.net/challenges/challenge-cd6d19866c42e274cd09604adaf4077b.tar.gz[/code] 
20多个key。估计是p, q重复使用的问题。

从每个key文件中提取N, e。 对每两个N求gcd，得到p(或q)。可求另一个 q= N/p 

之后求fi = (p - 1) * (q - 1)。 de = 1 in fi, 可以求 d.

知道N, e, d, p, q就可以解密了.
#!/usr/bin/python
import sys
import gmpy2
from gmpy2 import mpz
import os
import re

from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_OAEP

ns = [0] * 24
es = [0] * 24

for kf in os.listdir('./challenge'):
match = re.search(r"(\d+).key", kf)
if not match:
continue
f = open(os.path.join('./challenge', kf), 'r')
key = RSA.importKey(f.read())
i = int(match.group(1))
ns[i] = mpz(key.n)
es[i] = mpz(key.e)
f.close()

ps = [0] * len(ns)
qs = [0] * len(ns)
print len(ns)

for i in range(0, len(ns)):
for j in range(i + 1, len(ns)):
p = gmpy2.gcd(ns[i], ns[j])
if p != 1 and gmpy2.is_prime(p):
ps[i] = p
ps[j] = p

qs[i] = gmpy2.t_div(ns[i], p)
assert gmpy2.is_prime(qs[i]) == True

qs[j] = gmpy2.t_div(ns[j], p)
assert gmpy2.is_prime(qs[j]) == True

fis = [0] * len(ns)
ds = [0] * len(ns)
for i in range(0, len(ns)):
fis[i] = (ps[i] - 1) * (qs[i] - 1)
ds[i] = gmpy2.invert(es[i], fis[i])

pt = [0] * 24
for kf in os.listdir('./challenge'):
match = re.search(r"(\d+).enc", kf)
if not match:
continue
i = int(match.group(1))
f = open(os.path.join('./challenge', kf), 'r')
key = (long(ns[i]), long(es[i]), long(ds[i]), long(ps[i]), long(qs[i]))
key = RSA.construct(key)
cipher = PKCS1_OAEP.new(key)
pt[i] = cipher.decrypt(f.read())
f.close()
print ''.join(pt)