RSA

发表于

精彩，太精彩了

image-20240602220010889

enc文件不能直接打开，要通过脚本打开，大概率是内容有被加密

pub.key一看就是公钥，将公钥导入

再通过工具分解n为p和q

image-20240602220923919

之后就可以全部解出来了

然后最后我们打开enc文件需要的脚本如下

from Crypto.Util.number import bytes_to_long
with open("D:\\chorme的废物\\0eaf8d6c-3fe5-4549-9e81-94ac42535e7b (1)\\flag.enc","rb") as f:  #以二进制读模式，读取密文
    f = f.read()
    print(bytes_to_long(f))
    #c=29666689760194689065394649908301285751747553295673979512822807815563732622178

from libnum import n2s,s2n
from gmpy2 import *
e= 65537
n= 86934482296048119190666062003494800588905656017203025617216654058378322103517
p= 285960468890451637935629440372639283459
q= 304008741604601924494328155975272418463
d= 81176168860169991027846870170527607562179635470395365333547868786951080991441
c=29666689760194689065394649908301285751747553295673979512822807815563732622178
m=pow(c,d,n)
print(n2s(int(m)))
#flag{decrypt_256}