blocks|key|433158|text|对于浮点m，只要你的标准库实现写得很好，那么pow就是高效的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|433159|433160|也就是说，如果对该例程的调用占据了总运行时间的很大一部分。|433161|entityMap^0|M|3|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@$9|L|A|M|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|N|8|@]|D|@]|E|$]]|$1|G|3|H|5|6|7|O|8|@]|D|@]|E|$]]|$1|I|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|J|$]]

For floating point m, so long as your standard library implementation is well written, then <code>pow</code> will be efficient.

If m is an integer, and you hinted that it is, then you could use an array of pre calculated values.

You should only be worrying about this kind of thing if that routine is a bottleneck in your code. That is if the calls to that routine take a significant proportion of the total running time.

blocks|key|3457052|text|十在二进制机器上不是一个特殊的值，只有两个是。使用pow或exponentiation+by+squaring。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3457053|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/5625431/efficient-way-to-compute-pq-exponentiation-where-q-is-an-integer/5625446#5625446^0|P|3|T|Q|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

Ten is not a special value on a binary machine, only two is. Use <code>pow</code> or <a href="https://stackoverflow.com/questions/5625431/efficient-way-to-compute-pq-exponentiation-where-q-is-an-integer/5625446#5625446">exponentiation by squaring</a>.

blocks|key|3457070|text|不幸的是，使用IEEE754浮点表示法没有快速有效的方法来计算它。获得结果的最快方法是为您关心的每个m值构建一个表，然后执行查找。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3457071|entityMap^0|1E|1|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Unfortunately there is no fast and efficient way to calculate it using IEEE 754 floating point representation. The fastest way to get the result is to build a table for every value of <code>m</code> that you care about, and then just perform a lookup.

blocks|key|2348061|text|如果有一种快速有效的方法可以做到这一点，那么我相信你的CPU支持它，除非你运行在嵌入式系统上，在这种情况下，我希望电源(...)实现写得很好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2348062|10对我们来说很特别，因为我们大多数人都有10个手指。计算机只有两位数，所以2对它们来说是特别的。:)|2348063|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

If there's a fast and efficient way to do it then I'm sure your CPU supports it, unless you're running on an embedded system in which case I'd hope that the pow(...) implementation is well written.

10 is special to us as most of us have ten fingers. Computers only have two digits, so 2 is special to them. :)

blocks|key|5569004|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5569005|entityMap^0|0^^$0|@$1|2|3|-4|4|5|6|C|7|@]|8|@]|9|$]]|$1|A|3|-4|4|5|6|D|7|@]|8|@]|9|$]]]|B|$]]

Use lookup table there cant be more than 1000 floats and especially if m is integer.

blocks|key|3457204|text|IEEE754规定了一系列浮点格式。那些被广泛使用的是二进制的，这意味着基数10没有任何特殊之处。这与你认为"10也是一个特殊基数“的假设相反。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3457205|有趣的是，IEEE754-2008确实添加了十进制浮点格式(decimal32和朋友)。然而，我还没有遇到过它们的硬件实现。|offset|length|style|CODE|3457206|在任何情况下，您都不应该在分析代码并确定这确实是瓶颈之前对代码进行微优化。|3457207|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Decimal32_floating-point_format^0|0|U|9|U|9|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@$D|V|E|W|1|X]]|A|$]]|$1|H|3|I|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

IEEE 754 specifies a bunch of floating-point formats. Those that are in widespread use are binary, which means that base 10 isn't in any way special. This is contrary to your assumption that "10 is also a special base".

Interestingly, IEEE 754-2008 does add decimal floating-point formats (<a href="http://en.wikipedia.org/wiki/Decimal32_floating-point_format" rel="nofollow"><code>decimal32</code></a> and friends). However, I'm yet to come across hardware implementations of those.

In any case, you shouldn't be micro-optimizing your code before you've profiled it and established that this is indeed the bottleneck.

blocks|key|5569072|text|如果您可以在很长一段时间内使用log而不是n进行操作，则可以节省时间，因为|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5569073|n+=+pow(10*n,-m)|code-block|syntax|javascript|5569074|现在必须计算(使用定义l=+log10(n))|5569075|l+=+-m*(l%2B1)|5569076|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you could operate with log n instead of n for a significant time, you could save time because instead of

<pre><code>n = pow(10*n,-m)
</code></pre>

you now have to calculate (using the definition l = log10(n))

<pre><code>l = -m*(l+1)
</code></pre>

blocks|key|5569117|text|只是一些更多的想法，可能会引导你找到进一步的解决方案……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5569118|5569119|如果您对算法级别的优化感兴趣，您可以寻找一种并行方法。|ordered-list-item|5569120|您可以在系统/体系结构级别上加速使用Ipp+(用于英特尔处理器)，或者使用用于AMD+|5569121|的AMD核心数学库(ACML)来使用显卡的能力可能是另一种方式(例如用于NVIDEA卡的CUDA+)|5569122|我认为也值得看看OpenCL|5569123|<代码>G29|5569124|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|R|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|S|8|@]|9|@]|A|$]]|$1|H|3|I|5|E|7|T|8|@]|9|@]|A|$]]|$1|J|3|K|5|E|7|U|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|V|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|O|$]]

Just some more ideas which may lead you to further solutions... 

<ol>
<li>If you are interested in
optimization on algorithm level you
might look for a parallelized
approach.</li>
<li>You may speed up on
system/archtectural level on using Ipp
(for Intel Processors), or e.g. AMD
Core Math Library (ACML) for AMD</li>
<li>To use the power of your graphics
card may be another way (e.g. CUDA for NVIDEA cards)</li>
<li>I think it's also worth to look at
OpenCL</li>
</ol>

I want to compute 10 raised to the power minus <code>m</code>. In addition to use the math function <code>pow(10, -m)</code>, is there any fast and efficient way to do that?

What I ask such a simple question to the c++ gurus from SO is that, as you know, just like base 2, 10 is also a special base. If some value <code>n</code> times the 10's power minus <code>m</code>, it is equivalent to move n's decimal point to the left m times. I think it must be a fast and efficient way to cope with.

Fast way to compute n times 10 raised to the power of minus m

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我想计算10的幂，减去m。除了使用数学函数pow(10, -m)之外，还有什么快速有效的方法可以做到这一点吗？我问c++大师这么简单的问题是，正如你所知道的，就像基数2，10也是一个特殊的基数。如果某个值n乘以10的幂减去m，则相当于将n的小数点左移m次。我认为这一定是一种快速有效的应对方式。

问快速计算n乘以10的负m次幂
EN

回答 8

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问快速计算n乘以10的负m次幂EN

回答 8

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问快速计算n乘以10的负m次幂
EN