问题

你好，我对大O表示法不太了解，对于以下内容我有困惑，如果有人能够友好地解释一下如何计算，我将不胜感激。

int sum=1;
for(int count=n; count>0; count/=2) {
sum=sum*count;
}

<details>
<summary>英文:</summary>

Hi I am new to Big O Notation and having trouble specifying big O for the following if someone could kindly explain to me how to work it out please

int sum=1;
for(int count=n; count>0; count/=2) {
sum=sum*count;
}

</details>


# 答案1
**得分**: 3

每当`count`被2除，它将是`Theta(log n)`（从`n`到`0`）。因此，它也是`O(log(n))`。

<details>
<summary>英文:</summary>

As each time `count` is divided by 2, it will be `Theta(log n)` (from `n` to `0`). Hence, it is in `O(log(n))` as well.

</details>



# 答案2
**得分**: 1

在第一行中，代码将运行一次，因此对于它，复杂性为O(1)。
在第二行中，一个for循环将运行直到count > 0，我们可以看到它每次循环都会除以2，因此它将无限继续运行。因此，基本上无法对无限循环应用大O表示法。同样，您也不能将大O应用于循环内的代码。因此，如果存在任何O(infinite)，那么它可能是一个答案，但实际上不存在。因此，只有当n<=0时，才会执行代码，并且复杂性为O(1)。

<details>
<summary>英文:</summary>

In the first line, the code will run once, so for it, the complexity is O(1).
In second line a for loop will run until count &gt; 0 and we can see that it will be divided by 2 every time loop runs and so it will just go on continuously infinite times. So basically, you cannot apply Big O notation on an infinite loop. And so also you cannot apply big O to the code inside the loop as well. So if there exist any O(infinite) then it could be an answer, but it won&#39;t exist. So you cannot define time complexity until and unless n&lt;=0. if n&lt;=0 then only the code will run once and complexity will be O(1).

</details>



# 答案3
**得分**: 1

*count* 在每次迭代中减半。如果 `n = 64`：

step1 => count = 64
step2 => count = 32
step3 => count = 16
...

因此，其最坏情况的时间复杂度为 `O(log n)`。

然而，在这种情况下，您的循环体具有恒定数量的操作，因此，最佳情况、最坏情况和平均情况相同，并且：

    Ω(log n) = Θ(log n) = O(log n)

<details>
<summary>英文:</summary>

*count* gets halved per each iteration. If `n = 64`:

step1 => count = 64
step2 => count = 32
step3 => count = 16
...

Therefore, its worst case scenario has a `O(log n)` time complexity.

In this case, however, your loop's body has a constant number of operations, therefore, best, worst, and average case scenarios are same, and:

    Ω(log n) = Θ(log n) = O(log n)

</details>