2018百越杯Crypto RSA

题目给的是标准的RSA加密格式

#!/usr/bin/env python3
import gmpy2
from Crypto.Util.number import getPrime
from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_v1_5
from base64 import b64encode


flag = open('flag', 'r').read().strip() * 23


def encrypt(p, q, e, msg):
    while True:
        n = p * q
        try:
            phi = (p - 1)*(q - 1)
            pubkey = RSA.construct((int(n), int(e)))
            key = PKCS1_v1_5.new(pubkey)
            enc = b64encode(key.encrypt(msg))
            return enc
        except:
            p = gmpy2.next_prime(p**2 + q**2)
            q = gmpy2.next_prime(2*p*q)
            e = gmpy2.next_prime(e**2)


p = getPrime(128)
q = getPrime(128)
n = p*q
e = getPrime(64)
pubkey = RSA.construct((n, e))
with open('pubkey.pem', 'wb') as f:
    f.write(pubkey.exportKey())
with open('flag.enc', 'wb') as g:
    g.write(encrypt(p, q, e, flag.encode()))

这里掌握如何从标准公钥 密文中提取处理数据
exp:

import gmpy2
import libnum
import base64
from Crypto.PublicKey import RSA
from Crypto.Util.number import getPrime
from Crypto.Cipher import PKCS1_v1_5

pub = RSA.importKey(open(r'D:\浏览器下载\182_57791b62232b7838fc3eaec054aa9e1ce9b75b97\pubkey.pem').read())
c = open(r'D:\浏览器下载\182_57791b62232b7838fc3eaec054aa9e1ce9b75b97\flag.enc').read()
flag_decode = base64.b64decode(c)
n = pub.n
e = pub.e
#n=62078208638445817213739226854534031566665495569130972218813975279479576033261
#e=9850747023606211927
#此题注意p、q的位置不能交换，否则报错
q = 184333227921154992916659782580114145999
p = 336771668019607304680919844592337860739

def decrypt(p,q,e,msg):
	while True:
		n = p * q
		try:
			phi = (p - 1)*(q - 1)
			d = libnum.invmod(e,phi)
			private_key = RSA.construct((int(n), int(e), int(d), int(p), int(q)))
			decipher = PKCS1_v1_5.new(private_key)
			flag = decipher.decrypt(msg,None)
			return flag
			break
		except:
			p = gmpy2.next_prime(p**2 + q**2)
			q = gmpy2.next_prime(2*p*q)
			e = gmpy2.next_prime(e**2)

print (decrypt(p,q,e,flag_decode))