buuctf:rsa

最新推荐文章于 2025-05-30 22:55:59 发布

原创最新推荐文章于 2025-05-30 22:55:59 发布 · 804 阅读

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#服务器 #linux #运维

题目：含有flag.enc和pub.key两个文件

解题：用openssl 读取pub.key里的公钥，

openssl rsa -pubin -inform PEM -text < '/root/桌面/pub.key'

获得m=c0:33:2c:5c:64:ae:47:18:2f:6c:1c:87:6d:42:33:69:10:54:5a:58:f7:ee:fe:fc:0b:ca:af:5a:f3:41:cc:dd

转换为十进制：

a = 'c0:33:2c:5c:64:ae:47:18:2f:6c:1c:87:6d:42:33:69:10:54:5a:58:f7:ee:fe:fc:0b:ca:af:5a:f3:41:cc:dd'
b = a.replace(':','')
b
'c0332c5c64ae47182f6c1c876d42336910545a58f7eefefc0bcaaf5af341ccdd'
c = int(b,16)
c
86934482296048119190666062003494800588905656017203025617216654058378322103517

用factordb分解n，获得p、q

factordb 86934482296048119190666062003494800588905656017203025617216654058378322103517

安装factordb

pip install factordb-pycli

读取flag.enc文件的内容，转换为十六进制，后转为十进制；

import binascii
with open('/root/桌面/flag.enc','rb+') as f:
    f = f.read()
    a = binascii.hexlify(f)
    b = int(a,16)
    print(b)

再用脚本解密，但之前由于文字转换出现问题，以为自己数据出现问题，搜索相关WP，发现运用RSA库进行解密(不需要将c转换为数)：

import rsa
import gmpy2
c = 1854183526100811878807183372982532818560316522978821358738967769534081571682
p = 285960468890451637935629440372639283459
q = 304008741604601924494328155975272418463
e = 65537
n = 86934482296048119190666062003494800588905656017203025617216654058378322103517
fn = (p-1)*(q-1)

d = int(gmpy2.invert(e,fn))
key = rsa.PrivateKey(n,e,d,q,p)
with open(r'c:\111\flag.enc','rb') as f:    #文件路径
    f = f.read()
    print(rsa.decrypt(f,key))
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版权声明：本文为优快云博主「gal2xy」的原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.youkuaiyun.com/qq_52193383/article/details/119428246

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

#在pkcs标准中,pkcs#1规定,私钥包含(n,e,d,p,q)，相当于pow()

print(rsa.decrypt(f,key))

# f:公钥加密结果 key:私钥，相当于libnum.n2s(m)

之后通过其他方法，添加一行

decoded_message = libnum.n2s(m).decode('utf-8', 'replace')

#将解密后的整数 m 转换为字符串，并通过 'replace' 参数指定在遇到无法解码的字节时使用问号替代。

最终代码：

import libnum

p = 285960468890451637935629440372639283459
q = 304008741604601924494328155975272418463
e = 65537
n = 86934482296048119190666062003494800588905656017203025617216654058378322103517
phi = (p-1)*(q-1)
d = libnum.invmod(e, phi)
c = 29666689760194689065394649908301285751747553295673979512822807815563732622178
m = pow(c, d, n)

decoded_message = libnum.n2s(m).decode('utf-8', 'replace')
print(decoded_message)

flag{decrypt_256}