你的二次编码有错误的

int CGenerateWinCombination::calc_key(const MulUnitVector& allVec)
{
	int nResult = 0;
	int nLen = -1;

MulUnitVector::const_iterator iter = allVec.begin();
	while (iter != allVec.end())
	{
		BaseUnitVector::const_iterator secIter = iter->begin();
		while (secIter != iter->end())
		{
			nLen++;
			switch (*secIter)
			{
			case 2:
				nResult |= 0x3 << nLen;
				nLen += 2;
				break;
			case 3:
				nResult |= 0xF << nLen;
				nLen += 4;
				break;
			case 4:
				nResult |= 0x3F << nLen;
				nLen += 6;
				break;
			}
			secIter++;
		}
		nResult |= 0x1 << nLen;
		nLen++;
		iter++;
	}

return nResult;
}
在这个函数中，当出现牌张数大于1的时候，你不应该执行nResult |= 0x1 << nLen；操作，不然你的编码就不再是以0作为分割了

int CCheckWin::Calc_key(int nCardCount[34])
{
	int p = -1;
	int x = 0;
	bool b = false; // 首先是0

for (int i = 0; i < 27; i++)
	{
		if (nCardCount[i] != 0)
		{
			p++;
			b = true;
			switch (nCardCount[i])
			{
			case 2:
				x |= 0x3 << p;
				p += 2;
				break;
			case 3:
				x |= 0xF << p;
				p += 4;
				break;
			case 4:
				x |= 0x3F << p;
				p += 6;
				break;
			}

if (b && (nCardCount[i] == 0 || (i % 9 == 0)))
			{
				b = false;
				x |= 0x1 << p;
				p++;
			}
		}
	}
你这里的程序会计算的key 值和 生成函数的key值对应不上

根据程序【111】【111】【111】【2】【111】 生成的key是 291908（10进制）
而对于int cards[34] = {
	1,1,1,0,0,1,1,1,0,
	1,1,1,0,2,0,1,1,1,
	0,0,0,0,0,0,0,0,0,
	0,0,0,0,0,0,0
	}; 无赖子的key 是 6273（十进制）

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