[BUUCTF]SimpleRev

发表于 更新于 分类于 本文字数： 565 阅读时长 ≈ 2 分钟

不是PE文件, 直接拖进IDA分析, F5查看伪代码

image-20220122141058186 
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int __cdecl __noreturn main(int argc, const char **argv, const char **envp)
{
  int v3; // eax
  char v4; // [rsp+Fh] [rbp-1h]

  while ( 1 )
  {
    while ( 1 )
    {
      printf("Welcome to CTF game!\nPlease input d/D to start or input q/Q to quit this program: ");
      v4 = getchar();
      if ( v4 != 100 && v4 != 68 )
        break;
      Decry();	//重点就在这个函数里面
    }
    if ( v4 == 113 || v4 == 81 )
      Exit("Welcome to CTF game!\nPlease input d/D to start or input q/Q to quit this program: ", argv);		//退出函数这里为啥要传参进去?不是很懂
    puts("Input fault format!");
    v3 = getchar();
    putchar(v3);
  }
}

Decry()函数长这样

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unsigned __int64 Decry()
{
  char v1; // [rsp+Fh] [rbp-51h]
  int v2; // [rsp+10h] [rbp-50h]
  int v3; // [rsp+14h] [rbp-4Ch]
  int i; // [rsp+18h] [rbp-48h]
  int v5; // [rsp+1Ch] [rbp-44h]
  char src[8]; // [rsp+20h] [rbp-40h] BYREF
  __int64 v7; // [rsp+28h] [rbp-38h]
  int v8; // [rsp+30h] [rbp-30h]
  __int64 v9[2]; // [rsp+40h] [rbp-20h] BYREF
  int v10; // [rsp+50h] [rbp-10h]
  unsigned __int64 v11; // [rsp+58h] [rbp-8h]

  v11 = __readfsqword(0x28u);
  /*__readfsqword()函数从FS:[Offset]读取4字长的内容,同理还有__readfsbyte()等函数, 此处作用未知*/
  *(_QWORD *)src = 0x534C43444ELL;
  v7 = 0LL;
  v8 = 0;
  v9[0] = 0x776F646168LL;
  v9[1] = 0LL;
  v10 = 0;
  text = (char *)join(key3, v9);	//自定义的字符拼接,text="killshadow"
  strcpy(key, key1);
  strcat(key, src);		//key="ADSFKNDCLS"
  v2 = 0;
  v3 = 0;
  getchar();
  v5 = strlen(key);
  for ( i = 0; i < v5; ++i )
  {
    if ( key[v3 % v5] > 64 && key[v3 % v5] <= 90 )
      key[i] = key[v3 % v5] + 32;	//将key中每个字符由大写转成小写
    ++v3;
  }
  printf("Please input your flag:");
  while ( 1 )
  {
    v1 = getchar();
    if ( v1 == 10 )
      break;
    if ( v1 == 32 )
    {
      ++v2;
    }
    else
    {
      if ( v1 <= 96 || v1 > 122 )
      {
        if ( v1 > 64 && v1 <= 90 )
        {
          str2[v2] = (v1 - 39 - key[v3 % v5] + 97) % 26 + 97;
          ++v3;
        }
      }
      else
      {
        str2[v2] = (v1 - 39 - key[v3 % v5] + 97) % 26 + 97;
        ++v3;
      }
      if ( !(v3 % v5) )
        putchar(32);
      ++v2;
    }
  }
  if ( !strcmp(text, str2) )
    puts("Congratulation!\n");
  else
    puts("Try again!\n");
  return __readfsqword(0x28u) ^ v11;
}

主要就是逆向这一段, 得出v1

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while ( 1 )
  {
    v1 = getchar();
    if ( v1 == 10 )
      break;
    if ( v1 == 32 )
    {
      ++v2;
    }
    else
    {
      if ( v1 <= 96 || v1 > 122 )
      {
        if ( v1 > 64 && v1 <= 90 )
        {
          str2[v2] = (v1 - 39 - key[v3 % v5] + 97) % 26 + 97;
          ++v3;
        }
      }
      else
      {
        str2[v2] = (v1 - 39 - key[v3 % v5] + 97) % 26 + 97;
        ++v3;
      }
      if ( !(v3 % v5) )
        putchar(32);
      ++v2;
    }
  }

写出解密脚本

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#include <cstdio>
#include <cstring>

using namespace std;

int main()
{
    char key[16] = "adsfkndcls";
    char ciphertext[16] = "killshadow";
    char plaintext[16] = { 0 };
    char key_offset = 0;
    for (int i = 0; i < strlen(ciphertext); i++)
        for (char c = 32; c < 127; c++) //穷举所有可打印ASCII字符
        {
            char temp = c;
            if (('a' <= c && c <= 'z') or ('A' <= c && c <= 'Z'))
                temp = (c - 39 - key[key_offset % strlen(key)] + 97) % 26 + 97;
            if (temp == ciphertext[i])
            {
                plaintext[i] = c;
                key_offset++;
                break;
            }
        }
    printf("%s", plaintext);
}
//KLDQCUDFZO

所以flag就是flag{KLDQCUDFZO}