BUUCTF Reverse/[ACTF新生赛2020]Oruga

先看文件信息：没有加壳

IDA64位打开，分析代码

__int64 __fastcall main(int a1, char **a2, char **a3)
{
  __int64 result; // rax
  int i; // [rsp+0h] [rbp-40h]
  char s1[6]; // [rsp+4h] [rbp-3Ch] BYREF
  char s2[6]; // [rsp+Ah] [rbp-36h] BYREF
  char s[40]; // [rsp+10h] [rbp-30h] BYREF
  unsigned __int64 v8; // [rsp+38h] [rbp-8h]

  v8 = __readfsqword(0x28u);
  memset(s, 0, 25uLL);
  printf("Tell me the flag:");
  scanf("%s", s);
  strcpy(s2, "actf{");
  for ( i = 0; i <= 4; ++i )
    s1[i] = s[i];
  s1[5] = 0;
  if ( !strcmp(s1, s2) )
  {
    if ( sub_78A((__int64)s) )
      printf("That's True Flag!");
    else
      printf("don't stop trying...");
    result = 0LL;
  }
  else
  {
    printf("Format false!");
    result = 0LL;
  }
  return result;
}
           

重点是if语句中的条件

if ( sub_78A((__int64)s) )

，跟进查看

_BOOL8 __fastcall sub_78A(__int64 a1)
{
  int v2; // [rsp+Ch] [rbp-Ch]
  int v3; // [rsp+10h] [rbp-8h]
  int v4; // [rsp+14h] [rbp-4h]

  v2 = 0;
  v3 = 5;
  v4 = 0;
  while ( byte_201020[v2] != 0x21 )
  {
    v2 -= v4;
    if ( *(_BYTE *)(v3 + a1) != 'W' || v4 == -16 )
    {
      if ( *(_BYTE *)(v3 + a1) != 'E' || v4 == 1 )
      {
        if ( *(_BYTE *)(v3 + a1) != 'M' || v4 == 16 )
        {
          if ( *(_BYTE *)(v3 + a1) != 'J' || v4 == -1 )
            return 0LL;
          v4 = -1;
        }
        else
        {
          v4 = 16;
        }
      }
      else
      {
        v4 = 1;
      }
    }
    else
    {
      v4 = -16;
    }
    ++v3;
    while ( !byte_201020[v2] )
    {
      if ( v4 == -1 && (v2 & 0xF) == 0 )
        return 0LL;
      if ( v4 == 1 && v2 % 16 == 15 )
        return 0LL;
      if ( v4 == 16 && (unsigned int)(v2 - 240) <= 0xF )
        return 0LL;
      if ( v4 == -16 && (unsigned int)(v2 + 15) <= 0x1E )
        return 0LL;
      v2 += v4;
    }
  }
  return *(_BYTE *)(v3 + a1) == '}';
}
           

跟进byte_201020，看到老长一串，转到hex表

转成10进制

把空格替换为逗号

得到

0,0,0,0,35,0,0,0, 0,0,0,0,35,35,35,35,
0,0,0,35,35,0,0,0,79,79,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,79,79,0,80,80,0,0,0,
0,0,0,76,0,79,79,0,79,79,0,80,80,0,0,0,
0,0,0,76,0,79,79,0,79,79,0,80,0,0,0,0,
0,0,76,76,0,79,79,0, 0,0,0,80,0,0,0,0,
0,0,0,0,0,79,79,0, 0,0,0,80,0,0,0,0,
35,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0, 0,0,0,0,35,0,0,0,
0,0,0,0,0,0,77,77,77,0,0,0,35,0,0,0,
0,0,0,0,0,0,0,77,77,77,0,0,0,0,69,69,
0,0,0,48,0,77,0,77, 0,77,0,0,0,0,69,0,
0,0,0,0,0,0,0,0, 0,0,0,0,0,0,69,69,
84,84,84,73,0,77,0,77, 0,77,0,0,0,0,69,0,
0,84,0,73,0,77,0,77, 0,77,0,0,0,0,69,0,
0,84,0,73,0,77,0,77, 0,77,33,0,0,0,69,69,
           

然后我看代码看了半天，半天没有思路，，，看了大佬的wp才知道这玩意原来是个迷宫。。。

写个脚本输出一下

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int v2 = 0;
int v4 = 0;


int main()
{
    int i,j;
    char maze[] = {0,0,0,0,35,0,0,0, 0,0,0,0,35,35,35,35,
              0,0,0,35,35,0,0,0,79,79,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,79,79,0,80,80,0,0,0,
0,0,0,76,0,79,79,0,79,79,0,80,80,0,0,0,
0,0,0,76,0,79,79,0,79,79,0,80,0,0,0,0,
0,0,76,76,0,79,79,0, 0,0,0,80,0,0,0,0,
0,0,0,0,0,79,79,0, 0,0,0,80,0,0,0,0,
35,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0, 0,0,0,0,35,0,0,0,
0,0,0,0,0,0,77,77,77,0,0,0,35,0,0,0,
0,0,0,0,0,0,0,77,77,77,0,0,0,0,69,69,
0,0,0,48,0,77,0,77, 0,77,0,0,0,0,69,0,
0,0,0,0,0,0,0,0, 0,0,0,0,0,0,69,69,
84,84,84,73,0,77,0,77, 0,77,0,0,0,0,69,0,
0,84,0,73,0,77,0,77, 0,77,0,0,0,0,69,0,
0,84,0,73,0,77,0,77, 0,77,33,0,0,0,69,69};

   for(i = 0 ; i < 16 ;i++ )
   {

       for(j = 0 ; j < 16 ; j++)
       {
           if(maze[i * 16 + j] == 0)
              printf(" ");
           else if(maze[i * 16 + j] == 33)
              printf("#");
           else printf("*");
       }
       printf("\n");
   }

   return 0;
}

           

得到迷宫地图，起始点在左上角 I ，终点是#

I   *       ****
   **   **
        ** **
   * ** ** **
   * ** ** *
  ** **    *
     **    *
*
            *
      ***   *
       ***    **
   * * * *    *
              **
**** * * *    *
 * * * * *    *
 * * * * *#   **
           

然后这个在空格的位置可以随便走，但是一定要碰到障碍物才能停下来，像块肥皂一样。

根据这个，可以推出

M是下

W是上

J是左

E是右

while ( !byte_201020[v2] )       //在空格的地方一直走，直到撞到障碍物
    {
      if ( v4 == -1 && (v2 & 0xF) == 0 )
        return 0LL;
      if ( v4 == 1 && v2 % 16 == 15 )
        return 0LL;
      if ( v4 == 16 && (unsigned int)(v2 - 240) <= 0xF )    //加减16相当于上下，最终是以v2的值决定决定是否结束
        return 0LL;
      if ( v4 == -16 && (unsigned int)(v2 + 15) <= 0x1E )
        return 0LL;
      v2 += v4;
    }
           

然后就照着迷宫写出路径得到MEWEMEWJMEWJM

flag ： flag{MEWEMEWJMEWJM}