[ACTF新生赛2020]easyre

脱壳后查看main函数

int __cdecl main(int argc, const char **argv, const char **envp)
{
  _BYTE v4[12]; // [esp+12h] [ebp-2Eh] BYREF
  _DWORD v5[3]; // [esp+1Eh] [ebp-22h]
  _BYTE v6[5]; // [esp+2Ah] [ebp-16h] BYREF
  int v7; // [esp+2Fh] [ebp-11h]
  int v8; // [esp+33h] [ebp-Dh]
  int v9; // [esp+37h] [ebp-9h]
  char v10; // [esp+3Bh] [ebp-5h]
  int i; // [esp+3Ch] [ebp-4h]

  sub_401A10();
  qmemcpy(v4, "*F'\"N,\"(I?+@", sizeof(v4));
  printf("Please input:");
  scanf("%s", v6);
  if ( v6[0] != 'A' || v6[1] != 'C' || v6[2] != 'T' || v6[3] != 'F' || v6[4] != '{' || v10 != '}' )
    return 0;
  v5[0] = v7;
  v5[1] = v8;
  v5[2] = v9;
  for ( i = 0; i <= 11; ++i )
  {
    if ( v4[i] != byte_402000[*((char *)v5 + i) - 1] )
      return 0;
  }
  printf("You are correct!");
  return 0;
}

exp

s = [0x7E, 0x7D, 0x7C, 0x7B, 0x7A, 0x79, 0x78, 0x77, 0x76, 0x75,
     0x74, 0x73, 0x72, 0x71, 0x70, 0x6F, 0x6E, 0x6D, 0x6C, 0x6B,
     0x6A, 0x69, 0x68, 0x67, 0x66, 0x65, 0x64, 0x63, 0x62, 0x61,
     0x60, 0x5F, 0x5E, 0x5D, 0x5C, 0x5B, 0x5A, 0x59, 0x58, 0x57,
     0x56, 0x55, 0x54, 0x53, 0x52, 0x51, 0x50, 0x4F, 0x4E, 0x4D,
     0x4C, 0x4B, 0x4A, 0x49, 0x48, 0x47, 0x46, 0x45, 0x44, 0x43,
     0x42, 0x41, 0x40, 0x3F, 0x3E, 0x3D, 0x3C, 0x3B, 0x3A, 0x39,
     0x38, 0x37, 0x36, 0x35, 0x34, 0x33, 0x32, 0x31, 0x30, 0x2F,
     0x2E, 0x2D, 0x2C, 0x2B, 0x2A, 0x29, 0x28, 0x27, 0x26, 0x25,
     0x24, 0x23, 0x20, 0x21, 0x22]
key = [0x2a, 0x46, 0x27, 0x22, 0x4e, 0x2c, 0x22, 0x28, 0x49, 0x3f, 0x2b, 0x40]
flag = 'ACTF{'
for i in key:
    flag += chr(s.index(i) + 1)
print(flag+'}')