【Reverse】每日必逆0x02

BUU SimpleRev

附件 https://files.buuoj.cn/files/7458c5c0ce999ac491df13cf7a7ed9f1/SimpleRev

题解

• 查壳
  
  

• 拖入iad64进行静态分析
  
  关键伪代码

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);
  *(_QWORD *)src = 'SLCDN';
  v7 = 0LL;
  v8 = 0;
  v9[0] = 'wodah';
  v9[1] = 0LL;
  v10 = 0;
  text = (char *)join(key3, v9);                // text = "killshadow"
  strcpy(key, key1);                            // key = "ADSFK"
  strcat(key, src);                             // key = "ADSFKNDCLS"
  v2 = 0;
  v3 = 0;
  getchar();
  v5 = strlen(key);                             // v5 = 10
  for ( i = 0; i < v5; ++i )                    // Convert Capital to Lower-case
  {
    if ( key[v3 % v5] > 64 && key[v3 % v5] <= 90 )
      key[i] = key[v3 % v5] + 32;
    ++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;
}

• 程序流向
  
  将key3和v9连接并赋值给text,key变为"ADSFKNDCLS"（涉及小端存储，需要注意字符串的改变）=>转小写=>将输入的flag每个字母进行处理（(v1 - 39 - key[v3 % v5] + 97) % 26 + 97）=》将其与text进行比较

• 程序逆向
  
  获取text为"killshadow"，key="adsfkndcls"，再将算法进行逆向，便可以得出flag

• exp

key = "ADSFKNDCLS".lower()
text = "killshadow"
strs = "QWERTYUIOPASDFGHJKLZXCVBNM"
strs2 = "qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM"

flag = ""
for i in range(len(text)):
    for x in strs:
        if (ord(x)-39-ord(key[i%10])+97)%26+97 == ord(text[i]):
            flag+=x
print(flag)

也许你们会注意，这边字符串我有两种可能。应该是预期解都是大写吧，不然这题还有其他结果

总结

技术成长ing