ESA1024加密算法实现与解析

int __stdcall GetLastNotZero(LPDWORD lpData, DWORD nMax)
{
  int iIndex;
  iIndex = nMax;
  while(iIndex--)
  {
    if (lpData[iIndex])
      break;
  }
  return iIndex+1;
}

void memcpy_reverse(LPBYTE lpDest,LPBYTE lpSrc,DWORD dwSize)
{
  for(DWORD i=0;i<dwSize;i++)
    lpDest[i]=lpSrc[dwSize-i-1];
}
#define KEYSIZE 128
#define KEYCOUNT 32
int CalcJiaocha(DWORD Dword1, DWORD Dword2, LPDWORD Out1, LPDWORD Out2)
{
  unsigned int sumjiaocha; // eax@1
  unsigned int v5; // esi@1
  unsigned int high2; // esi@1
  int high2xlow1; // ecx@1
  unsigned int sumHighWord; // eax@3
  DWORD v9; // ecx@3
  int result; // eax@5
  
  high2 = Dword2 >> 16;

  high2xlow1 = high2 * (WORD)Dword1;
  *Out1 = (unsigned __int16)Dword2 * (unsigned __int16)Dword1;// 两个dword的低16位相乘
  sumjiaocha = high2xlow1 + (unsigned __int16)(Dword1 >> 16) * (unsigned __int16)Dword2;// 两个dword高低16位交叉相乘并相加
  *Out2 = high2 * (unsigned __int16)(Dword1 >> 16);// 两个dword的高16位相乘
  v5 = 0x10000u;
  if ( sumjiaocha >= high2xlow1 )
    v5 = 0;                                     // 如果加high2low1时没有溢出，则置v5=0;反之溢出则相当于高16位+1
  v9 = sumjiaocha << 16;
  *Out1 += sumjiaocha << 16;                    // out1加上"交叉乘积和"的低16位
  sumHighWord = sumjiaocha >> 16;
  if ( *Out1 < v9 )
    ++v5;                                       // 溢出则+1
  result = v5 + sumHighWord;
  *Out2 += result;                              // Out2加上交叉乘积的高16位，再加一个可能为0、1、0x10000、0x10001的变量
  return result;
}
int CalcBufferByNumAndKey(LPDWORD Key, LPDWORD lpBuf1End, DWORD KeyCount, DWORD dwValue)
{
  DWORD v7; // edx@5
  int v9; // ebx@5
  DWORD CalcValue2; // [sp+4h] [bp-14h]@5
  DWORD CalcValue1; // [sp+8h] [bp-10h]@5
  int v18= 0;
  if ( dwValue )
  {
    for(DWORD i=0;i<KeyCount;i++)
    {
      CalcJiaocha(dwValue,Key[i],&CalcValue1,&CalcValue2);
      v9=lpBuf1End[i]-v18;
      lpBuf1End[i]=v9-CalcValue1;