[FlareOn3]Challenge1 题解

_BYTE *__cdecl sub_401260(int a1, unsigned int a2)
{
  int v3; // [esp+Ch] [ebp-24h]
  int v4; // [esp+10h] [ebp-20h]
  int v5; // [esp+14h] [ebp-1Ch]
  int i; // [esp+1Ch] [ebp-14h]
  unsigned int v7; // [esp+20h] [ebp-10h]
  _BYTE *v8; // [esp+24h] [ebp-Ch]
  int v9; // [esp+28h] [ebp-8h]
  int v10; // [esp+28h] [ebp-8h]
  unsigned int v11; // [esp+2Ch] [ebp-4h]
  v8 = malloc(4 * ((a2 + 2) / 3) + 1);
  if ( !v8 )
    return 0;
  v11 = 0;
  v9 = 0;
  while ( v11 < a2 )
  {
    v5 = *(unsigned __int8 *)(v11 + a1);
    if ( ++v11 >= a2 )
    {
      v4 = 0;
    }
    else
    {
      v4 = *(unsigned __int8 *)(v11 + a1);
      ++v11;
    }
    if ( v11 >= a2 )
    {
      v3 = 0;
    }
    else
    {
      v3 = *(unsigned __int8 *)(v11 + a1);
      ++v11;
    }
    v7 = v3 + (v5 << 16) + (v4 << 8);
    v8[v9] = byte_413000[(v7 >> 18) & 0x3F];
    v10 = v9 + 1;
    v8[v10] = byte_413000[(v7 >> 12) & 0x3F];
    v8[++v10] = byte_413000[(v7 >> 6) & 0x3F];
    v8[++v10] = byte_413000[v3 & 0x3F];
    v9 = v10 + 1;
  }
  for ( i = 0; i < dword_413040[a2 % 3]; ++i )
    v8[4 * ((a2 + 2) / 3) - i - 1] = 61;
  v8[4 * ((a2 + 2) / 3)] = 0;
  return v8;
}

base64_data=[0x5a,0x59,0x58,0x41,0x42,0x43,0x44,0x45,0x46,0x47,0x48,0x49,0x4a,0x4b,0x4c,0x4d,0x4e,0x4f,0x50,0x51,0x52,0x53,0x54,0x55,0x56,0x57,0x7a,0x79,0x78,0x61,0x62,0x63,0x64,0x65,0x66,0x67,0x68,0x69,0x6a,0x6b,0x6c,0x6d,0x6e,0x6f,0x70,0x71,0x72,0x73,0x74,0x75,0x76,0x77,0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x2b,0x2f]
base64_table=''
for i in range(len(base64_data)):
    base64_table+=chr(base64_data[i])
print(base64_table)

import base64
str1 = "x2dtJEOmyjacxDemx2eczT5cVS9fVUGvWTuZWjuexjRqy24rV29q"
base64_table = "ZYXABCDEFGHIJKLMNOPQRSTUVWzyxabcdefghijklmnopqrstuvw0123456789+/"
base64_str = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
decoded_str = base64.b64decode(str1.translate(str.maketrans(base64_table, base64_str)))
print(decoded_str)