我正在寻找一种简单的算法，给出宽度w和高度h的矩形，分割矩形成n或多或少相等的尺寸和形状的矩形，并计算这些矩形的中心。

I'm looking for a simple algorithm that, given a rectangle with width w and height h, splits the rectangle into n more or less equal sized and shape rectangles and calculates the center of these rectangles.

编辑：忘了提，形状应尽可能类似于正方形

Forgot to mention that the shapes should be as similar as possible to a square.

任何提示如何开始？

推荐答案

一个简单的算法是垂直分割成的高度h和宽度w / N N相等大小的条。

A simple algorithm is to split vertically into n equal sized strips of height h and width w/n.

如果你假设初始矩形的角落（0,0）和（W，H），然后使用该算法的I 日矩形会对中心（W / N *（1 +½ ）中，h / 2），为0℃= I＆其中; ñ。

If you assume that the initial rectangle has corners (0,0) and (w,h) then using this algorithm the ith rectangle would have center (w / n * (i + ½), h/2), for 0 <= i < n.

更新：尝试发现的数量n成因子对（I，J）中的所有因式分解使得I * J = n，并且找到因子对，使得各因素的比率最接近边的比值矩形的。然后使用两种因素来创建更小的矩形规则的网格。

Update: try finding all the factorizations of the number n into factor pairs (i, j) such that i * j = n, and find the factor pair such that the ratio of the factors is closest to the ratio of the sides of the rectangle. Then use the two factors to create a regular grid of smaller rectangles.

例如当n是10，则可以在（1，10）选择，（2,5），（5,2），（10，1）。下面是使用因素的一个例子格（5,2）：

For example when n is 10, you can choose between (1, 10), (2, 5), (5, 2) and (10, 1). Here is an example grid using the factors (5, 2):

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|      |      |      |      |      |
|      |      |      |      |      |
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|      |      |      |      |      |
|      |      |      |      |      |
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如果您最初的矩形的宽60，高20，然后运用因子对（5,2）将十个矩形的大小（60/5，20/2）=（12，10），这是接近正方形。

If your initial rectangle has width 60 and height 20 then using the factor pair (5, 2) will give ten rectangles of size (60/5, 20/2) = (12, 10) which is close to square.