BUUCTF - re - rsa

BUUCTF - re - rsa

http://tool.chacuo.net/cryptrsakeyparse 将key文件拖进去解析

在这里插入图片描述

可以得到

e=65537

n=0xC0332C5C64AE47182F6C1C876D42336910545A58F7EEFEFC0BCAAF5AF341CCDD

十进制就是n=86934482296048119190666062003494800588905656017203025617216654058378322103517

如下网址得到p q

http://www.factordb.com/index.php?query=86934482296048119190666062003494800588905656017203025617216654058378322103517

p=285960468890451637935629440372639283459

q=304008741604601924494328155975272418463

然后就是常规的RSA,脚本直接得

import gmpy2
import rsa
e = 65537
n = 86934482296048119190666062003494800588905656017203025617216654058378322103517
p = 285960468890451637935629440372639283459
q = 304008741604601924494328155975272418463

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

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

with open("C:\\Users\\86177\\Desktop\\flag.txt", "rb+") as f:
    f = f.read()
    print(rsa.decrypt(f, key))
#flag{decrypt_256}

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转载自blog.csdn.net/yzl_007/article/details/121212488