For a web developer, it is very important to know how to design a web page's size. So, given a specific rectangular web page’s area, your job by now is to design a rectangular web page, whose length L and width W satisfy the following requirements:

1. The area of the rectangular web page you designed must equal to the given target area.

2. The width W should not be larger than the length L, which means L >= W.

3. The difference between length L and width W should be as small as possible.

You need to output the length L and the width W of the web page you designed in sequence.

Example:

Input: 4
Output: [2, 2]
Explanation: The target area is 4, and all the possible ways to construct it are [1,4], [2,2], [4,1]. 
But according to requirement 2, [1,4] is illegal; according to requirement 3,  [4,1] is not optimal compared to [2,2]. So the length L is 2, and the width W is 2.

Note:

1. The given area won't exceed 10,000,000 and is a positive integer
2. The web page's width and length you designed must be positive integers.

這道題讓我們根據面積來求出矩形的長和寬，要求長和寬的差距儘量的小，那麼就是說越接近正方形越好。那麼我們肯定是先來判斷一下是不是正方行，對面積開方，如果得到的不是整數，說明不是正方形。那麼我們取最近的一個整數，看此時能不能整除，如果不行，就自減1，再看能否整除。最壞的情況就是面積是質數，最後減到了1，那麼返回結果即可，參見程式碼如下：

解法一：

class Solution {
public:
    vector<int> constructRectangle(int area) {
        int r = sqrt(area);
        while (area % r != 0) --r;
        return {area / r, r};
    }
};

如果我們不想用開方運算sqrt的話，那就從1開始，看能不能整除，迴圈的終止條件是看平方值是否小於等於面積，參見程式碼如下：

解法二：

class Solution {
public:
    vector<int> constructRectangle(int area) {
        int r = 1;
        for (int i = 1; i * i <= area; ++i) {
            if (area % i == 0) r = i;
        }
        return {area / r, r};
    }
};

參考資料：