Question

简单的问题,是有可能简化(或以较低费用运作取代司或组合)

(k/m)%n

在变数为单位的情况下,操作者为C风格分工和模块操作者。

请允许我略微回答问题,除非变数是基数2,在什么条件下(例如某些变数可能保持不变)可以简化表述(或部分使用基2操作)以删除分部分或模块?

这对我来说是学习数字理论,特别是基础2的trick计,而不是在优化业绩方面进行。

谢谢。

Answer 1

各位可以采用这样的方法,对持小不变的提名。

k/m=k*(1/m)
x=(1<<16)/m
k/m=(k*x)>>16

答案可能并不准确,取决于投入。

关于与小奇常分母的划分,您可使用

3   2863311531      11  3123612579
5   3435973837      13  3303820997
7   3067833783      15  4008636143
9    954437177      17  4042322161

x/11 == x*3123612579 % 2^32

<代码>% 2^32 当然在32英寸内无。采用这种数字,甚至数字因素,然后适用。

x/44 == (x*3123612579 % 2^32) >> 2

Hackers Delight有一章涉及分类。

简单模块和权力分工。

x%m == x&(m-1)
x/m == x>>log2(m) // assumes log2(m) is known, not calculated

Answer 2

For arbitrary-precision integers, I recommendation look at

它是一种硬件方法,但提供了可加以调整的假象。

Answer 3

明显的优化:

m == 1 : The answer will just be k % m.
n == 1 : The answer is always 0.
m is a power of 2 : e.g. if m is 4, you can use (k >> 2) % n;
n is a power of 2 : expression becomes (k / m) & (n - 1);

第1号和2号的检查是微不足道的。

检查两种权力时,采用:

void isPowerOfTwo(unsigned int x)
{
    return x & (x - 1) == 0;
}

Answer 4

加上Peter Alexander的答复

页: 1 页: 1 页: 1

(1) k; 答案总是0。

2)k = m: 回答总是1(除非是1,见5)。

3)k / m < n: 答案为k/m。

4) k < ( m * n ): The answer is always k / m. This particular condition is not very conducive to optimization since m*n would not be reused and it should not be much faster than modulo unless m and/or n are powers of 2 in which case you d still be better using 7. and/or 8.

参考:

页: 1

7) m 功率为2:例如,如果面积为4,你可以使用(k >> 2)n;

8)n 是2 :表达(k /m)和(n - 1)

Answer 5

I suppose that it depends. A right arithmetic shift would give you division by 2 and with some additional arithmetic you could make that into a division by any number. Similarly I would think that you could do the same with the modulo operator. In reality, unless your processor is lacking the necessary hardware, I wouldn t have thought that you would gain anything.

edit Actually thinking about it, more thought is needed here as strictly speaking, for a negative number a signed shift wouldn t work as a divide by 2.

如果我正确地记住,在C标准中,算术转变权的行为没有界定(关于权力的会谈2),汇编者也依赖。

如果我们只想一下数字理论的逻辑,那就是一个不同的问题。让我思考一下。

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