整数ceil（sqrt（x）） [英] Integer ceil(sqrt(x))

问题描述

答案提供了以下用于计算楼层的代码（sqrt（x））只使用整数。是否可以使用/修改它来返回 ceil（sqrt（x））？或者，计算这样的价值的首选方法是什么？

The answer gives the following code for computing floor(sqrt(x)) using just integers. Is it possible to use/modify it to return ceil(sqrt(x)) instead? Alternatively, what is the preferred way to calculate such value?

编辑：谢谢大家到目前为止我道歉，我应该让它更明确：我希望那里使用 floor（sqrt（x）），可能加一个更自然的方式。 floor 版本使用Newton的方法从上面接近root，我认为可能从下面接近它或类似的就可以了。

Thank you all so far and I apologise, I should have make it more explicit: I was hoping there is more "natural" way of doing this that using floor(sqrt(x)), possibly plus one. The floor version uses Newton's method to approach the root from above, I thought that maybe approaching it from below or similar would do the trick.

例如，答案甚至提供了如何舍入到最接近的整数：只需输入 4 * x 算法。

For example the answer even provides how to round to nearest integer: just input 4*x to the algorithm.

推荐答案

如果 x 是一个精确的正方形，则平方根的上限和下限是相等的;否则，天花板比平方根多一个。所以你可以使用（在Python中），

If x is an exact square, the ceiling and the floor of the square root are equal; otherwise, the ceiling is one more than the square root. So you could use (in Python),

result = floorsqrt(x)
if result * result != x:
    result += 1

修改链接的代码并不是一个好主意，因为该代码使用Newton-Raphson方法计算平方根的一些属性。关于该方法已经开发了很多理论，并且代码使用了该理论。我展示的代码不像修改链接代码那样整洁，但它比更改代码更安全，也可能更快。

Modifying the code you linked to is not a good idea, since that code uses some properties of the Newton-Raphson method of calculating the square root. Much theory has been developed about that method, and the code uses that theory. The code I show is not as neat as modifying your linked code but it is safer and probably quicker than making a change in the code.

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