攻防世界 cr4-poor-rsa

题目

给了一个压缩包，内有文件：flag.b64（打开发现是base64编码的）和key.pub

解题过程

1. 首先利用Crypto.PublicKey的RSA模块从key.pub中获取公钥信息，这个RSA相关知识点中已有介绍
2. 得到n后可以拿到大数分解，得到

p = 863653476616376575308866344984576466644942572246900013156919
q = 965445304326998194798282228842484732438457170595999523426901

3. 由p,q,e得到d（RSA相关知识点中已有介绍）
4. rsa库的PrivateKey生成私钥
5. 打开flag.b64并用base64解码，再利用key解密即可得到flag

from Crypto.PublicKey import RSA
from gmpy2 import invert
import rsa
from base64 import b64decode

f = open("F:\ChromeCommon\key.pub","rb").read()
pub = RSA.importKey(f)
n = pub.n
e = pub.e

p = 863653476616376575308866344984576466644942572246900013156919
q = 965445304326998194798282228842484732438457170595999523426901
d = int(invert(e, (p-1)*(q-1)))

key = rsa.PrivateKey(n,e,d,p,q)
f = open("F:\ChromeCommon\flag.b64",'r').read()
c = b64decode(f)
flag = rsa.decrypt(c,key)
print(flag)

ALEXCTF{SMALL_PRIMES_ARE_BAD}