优化定点 sqrt

我做了我认为是一个很好的fixed-point square root algorithm：

template<int64_t M, int64_t P>
typename enable_if<M + P == 32, FixedPoint<M, P>>::type sqrt(FixedPoint<M, P> f)
{
    if (f.num == 0)
        return 0;
    //Reduce it to the 1/2 to 2 range (based around FixedPoint<2, 30> to avoid left/right shift branching)
    int64_t num{ f.num }, faux_half{ 1 << 29 };
    ptrdiff_t mag{ 0 };
    while (num < (faux_half)) {
        num <<= 2;
        ++mag;
    }
    int64_t res = (M % 2 == 0 ? SQRT_32_EVEN_LOOKUP : SQRT_32_ODD_LOOKUP)[(num >> (30 - 4)) - (1LL << 3)];
    res >>= M / 2 + mag - 1; //Finish making an excellent guess
    for (int i = 0; i < 2; ++i)
        //                                |   /
        //                                |  /
        //                              _| V L
        res = (res + (int64_t(f.num) << P) / res) >> 1; //Use Newton's method to improve greatly on guess
        //                               7 A r
        //                              /  |  
        //                             /   |   
        //                       The Infamous Time Eater
    return FixedPoint<M, P>(res, true);
}

但是，在分析（在发布模式下）之后，我发现这里的划分占用了该算法花费的 83% 的时间。我可以通过用乘法代替除法来加快 6 倍的速度，但这是错误的。不幸的是，我发现整数除法比乘法慢得多。有什么方法可以优化这一点吗？

如果此表是必需的。

const array<int32_t, 24> SQRT_32_EVEN_LOOKUP = {
     0x2d413ccd, //magic numbers calculated by taking 0.5 + 0.5 * i / 8 from i = 0 to 23, multiplying by 2^30, and converting to hex
     0x30000000,
     0x3298b076,
     0x3510e528,
     0x376cf5d1,
     0x39b05689,
     0x3bddd423,
     0x3df7bd63,
     0x40000000,
     0x41f83d9b,
     0x43e1db33,
     0x45be0cd2,
     0x478dde6e,
     0x49523ae4,
     0x4b0bf165,
     0x4cbbb9d6,
     0x4e623850,
     0x50000000,
     0x5195957c,
     0x532370b9,
     0x54a9fea7,
     0x5629a293,
     0x57a2b749,
     0x59159016
};

SQRT_32_ODD_LOOKUP只是SQRT_32_EVEN_LOOKUP除以sqrt(2)。