Whctf 2017 -UNTITLED- Writeup

Whctf 2017 -UNTITLED- Writeup

转载请表明出处http://www.cnblogs.com/WangAoBo/p/7541481.html

分析：

  • 下载下来的附件是一个py脚本，如下

     1 from Crypto.Util.number import getPrime,long_to_bytes,bytes_to_long
 2 import primefac
 3 import time
 4 from os import urandom
 5 import hashlib
 6 import sys
 7 class Unbuffered(object):
 8    def __init__(self, stream):
 9        self.stream = stream
10    def write(self, data):
11        self.stream.write(data)
12        self.stream.flush()
13    def __getattr__(self, attr):
14        return getattr(self.stream, attr)
15 import sys
16 sys.stdout = Unbuffered(sys.stdout)
17 def gen_args():
18     p=getPrime(1024)
19     q=getPrime(1024)
20     n=p*q
21     e=0x10001
22     d=primefac.modinv(e,(p-1)*(q-1))%((p-1)*(q-1))
23     return (p,q,e,n,d)
24 def proof():
25     salt=urandom(4)
26     print salt.encode("base64"),
27     proof=raw_input("show me your work: ")
28     if hashlib.md5(salt+proof.decode("base64")).hexdigest().startswith("0000"):
29         print "checked success"
30         return 1
31     return 0
32 
33 def run():
34     if not proof():
35         return
36     m=int(open("/home/bibi/PycharmProjects/work/whctf/flag","r").read().encode("hex"),16)#flag{*}
37     (p,q,e,n,d)=gen_args()
38     c=pow(m,e,n)
39     print "n:",hex(n)
40     print "e:",hex(e)
41     print "c:",hex(c)
42     t=int(hex(m)[2:][0:8],16)
43     u=pow(t,e,n)
44     print "u:",hex(u)
45     print "===="
46     x=int(hex(m)[2:][0:8]+raw_input("x: "),16)
47     print "===="
48     y=int(raw_input("y: "),16)
49     if (pow(x,e,n)==y and pow(y,d,n)==t):
50         print "s:",hex(int(bin(p)[2:][0:568],2))
51 run()

     

    ​

  • nc连上之后如下：

  • 结合py代码分析，首先要通过proof()函数的验证，即要满足：

    hashlib.md5(salt+proof.decode("base64")).hexdigest().startswith("0000")

    写一个脚本爆破，使盐值和输入内容的base64解码的md5开头四位为0000

    #!/usr/bin/env python
# -*- coding: utf-8 -*-
__Auther__ = 'M4x'

import hashlib
import string

#base64编码后的范围
dic = string.ascii_letters + string.digits + "+/"

#盐值
salt = '+/DlHw=='.decode('base64')

#先尝试爆破4位
for a in dic:
    for b in dic:
        for c in dic:
            for d in dic:
                proof = a + b + c + d
                try:
                    if hashlib.md5(salt + proof.decode("base64")).hexdigest().startswith("0000"):
                        print proof
                        exit(0)
                except:
                    pass

     

    很快就爆破出多组结果

    随便找一组提交，通过了proof函数的验证，得到了n, e, c, u

 

  • 再分析题目给的py脚本，可以看出

    • c即为flag经过RSA加密的密文，其中n和e已知

    • u为m的前8位（根据flag形式，即为f）经过RSA加密后的值

    • x=int(hex(m)[2:][0:8]+raw_input("x: "),16)

y=int(raw_input("y: "),16)

pow(x,e,n)==y and pow(y,d,n)==t

      以上三行连起来分析，很容易得出当我们输入的x为空，y为u时，即可通过最后一步if的验证，从而得到p的前568位（输入y时记得去掉最后的L）

  • 至此整理一下我们得到的信息：

    • flag加密后的密文c
    • 加密所用到的n和e
    • p的前568位

    很容易联想到恢复p，从而算出q，再解RSA就能拿到flag

步骤：

  • 这个时候想到了国赛的一道类似题目Partial，也是知道n，e和高位p，需要恢复p，因此也选用相同的方法Coppersmith Attack(https://github.com/Gao-Chuan/RSA-and-LLL-attacks#factoring-with-high-bits-known)，但Coppersmith Attack方法需要我们最少知道576位p，已知568位，差了两个16进制数，根据官方给的hint

    很明显我们需要爆破出要补上的两位

  • 在强大的sagemath上写了一个爆破的代码，在线运行（http://sagecell.sagemath.org/）

     1 n = 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
 2 p = 0xb447abcd768378f05675b98f4724e934b1a7251749b14b11d3af19d3a47e98dbf90b94a77a01ab76e6a7f99d5b79cfce8e9edfcc7b626ed0f1699d743fa78bd73ff4a03f904bde
 3 
 4 import string
 5 dic = string.digits + "abcdef"
 6 
 7 for a in dic:
 8     for b in dic:
 9         pp = hex(p) + a + b
10         #p需要用0补全到1024位
11         pp += '0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000'
12         #要加的数字与补全p时0的个数有关
13         pp = int(pp, 16)
14         p_fake = pp+0x10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
15         pbits = 1024
16         kbits = pbits-576
17         pbar = p_fake & (2^pbits-2^kbits)
18         print "upper %d bits (of %d bits) is given" % (pbits-kbits, pbits)
19         PR.<x> = PolynomialRing(Zmod(n))
20         f = x + pbar
21         try:
22             x0 = f.small_roots(X=2^kbits, beta=0.4)[0]  # find root < 2^kbits with factor >= n^0.4
23             print x0 + pbar
24         except:
25             pass

     

    爆破出了p如下：

  • 现在知道了n, e, c, p，解开RSA就可以了，python脚本如下（当然把解RSA的过程写在sage代码中也是可以的）

     1 #!/usr/bin/env python
 2 # -*- coding: utf-8 -*-
 3 __Auther__ = 'M4x'
 4 
 5 p = 126596896828983947657897211653294325357694173315986362964483543178327683872006349352506228192861938882562062524573153829867465009733178457399135420215887364009777012624212242069216745138202953735034716032666189414323613790242613717531697843979604409625853777348356827810939640137432278820298916431800157020739
 6 n = 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
 7 e = 0x10001
 8 c = 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
 9 
10 import libnum
11 import gmpy2
12 
13 q = n / p
14 assert n == p * q
15 
16 d = gmpy2.invert(e, (p - 1) * (q - 1))
17 m = pow(c, d, n)
18 print libnum.n2s(m)

     

    python 解RSA的姿势（http://www.cnblogs.com/WangAoBo/p/7513811.html）

    运行，即可得到flag

     

posted @ 2017-09-18 09:11  M4x  阅读(2786)  评论(0)  编辑  收藏  举报