BUUCTF RSA

給的文件夾中有個pub.key，里面是公鑰，

-----BEGIN PUBLIC KEY-----
MDwwDQYJKoZIhvcNAQEBBQADKwAwKAIhAMAzLFxkrkcYL2wch21CM2kQVFpY9+7+
/AvKr1rzQczdAgMBAAE=
-----END PUBLIC KEY-----

 在線分解

 

得到n，e

用http://www.factordb.com

 

 得到p和q，寫腳本

import gmpy2
import rsa

e=65537
n=86934482296048119190666062003494800588905656017203025617216654058378322103517
p=285960468890451637935629440372639283459
q=304008741604601924494328155975272418463

phin = (p-1) * (q-1)
d=gmpy2.invert(e, phin)

key=rsa.PrivateKey(n,e,int(d),p,q)

with open("flag.enc","rb") as f:
    f=f.read()
    print(rsa.decrypt(f,key))

 得到flag