如何找到一些与循环的因素是什么？ [英] How to find the factors of a number with for loops?

问题描述

嘿，我需要一些帮助，我的AP的计算机科学功课。在任务，我需要打印传递的变量，z的因素。这是我在我的方法，到目前为止得到

Hey I need some help with my AP Comp Sci homework. In the assignment I need to print the factors of a passed variable, z. This is what I've gotten so far in my method

public static void printFactors(int z) {
    for(int x=1; x<=z; x++) {
        if(z%x.......) {
            System.out.println(x);
        }
    }
}

我将如何完成呢？还是我即使是在正确的轨道上？谢谢！

How would I go about finishing this? Or am I even on the right track? Thanks!

推荐答案

至于你的问题，你想获得的所有的因素，我可能会认为，你想要的完整列表，但一个明显的因素整数。

As your question, you want to get ALL factors, I might think, you want the full list, but distinct factors of an integer.

所以，你只能遍历从1整数x为的sqrt（Z），但SQRT本身就是太慢了，你可以循环x直到 X * X＆GT; ž，见下面循环。

So, you can only iterates the integer x from 1 to sqrt(z), but sqrt itself is too slow, and you can loop x until x*x > z, see the for loop below.

更多，你应该注意，如果以Z％×== 0

More, you should take care that if z % x == 0

例如：Z = 18，X = 3，Z％X = 0

for example: z = 18, x =3, z % x = 0

则z / x必须也是一个因素，太！它应该是输出

then z / x must be also a factor, too! It should be output.

但要避免产生重复的因素，如果以Z = = X * X ，然后 X == Z / X ，这两个因素 X 和 Z / X 是重复的，它应该只被输出一次。

but to avoid duplicated factors produced, if z == x * x, then x == z / x, the two factors x and z/x is duplicate, it should only be output once.

这是一个经典算法问题，这个实现是普遍使用的一种，它与时间复杂度 O（开方（Z））。

This is a classical algorithm issue, and this implemetation is the generally used one, which with time complexity O(sqrt(z)).

见code：

public static void printFactors(int z) {
    for(int x=1; x * x <= z; x++) {
        if(z % x == 0) {
            System.out.println(x);
            if(x * x != z) System.out.println(z / x);
        }
    }
}

有一个尝试！

Have a try!

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