快速的方式找到一个数的倍数 [英] Fast way to find the next multiple of a number

问题描述

我需要找到从基数开始数的第一个倍数。例如：3 7的第一个多是9.我第一次尝试这样做：

I need to find the first multiple for a number starting from a base number. For example: The first multiple of 3 from 7 is 9. My first attempt was to do this:

multiple = baseNumber
while(multiple%number !=0 )
    multiple++

最后，多重将有数量后 baseNumber 第一多个。的问题是，当号变得过大，迭代的数量变得过多。所以我的问题是：有没有更快的方法来做到这一点。

At the end, "multiple" will have the first multiple of number after baseNumber. The problem is that when number becomes too large, the number of iterations becomes too many. So my question is: is there a faster way to do this?

推荐答案

尝试

multiple = baseNumber + number - 1;
mutliple -= (multiple % number);

这确实是在固定时间内。

That does it in constant time.

首先，我们添加 - 1 ，以确保我们有一个数字至少应为下​​一个倍数，但比后一小。然后，我们通过数量减去的余数，以确保我们有所需的倍数。

First, we add number - 1 to make sure that we have a number at least as large as the next multiple but smaller than the one after that. Then we subtract the remainder of the division by number to make sure we have the desired multiple.

注：以上的回报，因为原来的code呢， baseNumber 如果这已经是数的倍数。如果这是不希望的，删除 - 1 在第一行

Note: The above returns, as the original code does, baseNumber if that is already a multiple of number. If that isn't desired, remove the - 1 in the first line.

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