題目描述：

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.

要完成的函數：

vector<int> constructRectangle(int area)

說明：

1、這道題目不難，給定一個乘積，要求輸出兩個因子（大的在前，小的在後），兩個因子的差應該越小越好。

2、看到這些限定條件，我們首選就是開方，然後在開方得到的數值附近找到一個乘積能夠整除的數。

代碼如下：

    vector<int> constructRectangle(int area) 
    {
        int eu=ceil(sqrt(area));
        while(area%eu)
            eu+=1;
        return {eu,area/eu};
    }

上述代碼accepted，實測80ms，beats 27.85% of cpp submissions……

這麽低的嗎？但我看評論區的代碼采用的也是跟我一樣的方法，為什麽他們就能4ms……

3、改進：

細細對比了一下評論區的代碼，他們采用的是eu不斷地減一，而我用的是eu不斷地加一。

因為eu小的話比較容易除？比如2889除以11，跟2889除以3比起來，肯定後者比較好做除法。

所以如果采用eu不斷地減一的方法，的確能夠節省很多做除法的時間，而兩者的效果是一樣的。

代碼如下：

    vector<int> constructRectangle(int area) 
    {
        int eu=floor(sqrt(area));
        while(area%eu)
            eu-=1;
        return {area/eu,eu};
    }

實測3ms，beats 98.48% of cpp submissions。

leetcode-492-Construct the Rectangle