给的文件夹中有个pub.key,里面是公钥,
-----BEGIN PUBLIC KEY----- MDwwDQYJKoZIhvcNAQEBBQADKwAwKAIhAMAzLFxkrkcYL2wch21CM2kQVFpY9+7+ /AvKr1rzQczdAgMBAAE= -----END PUBLIC KEY-----
在线分解
得到n,e
得到p和q,写脚本
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
原文地址:https://www.cnblogs.com/harmonica11/p/11504291.html
时间: 2024-08-03 20:03:54