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