如何在 Java 中舍入到 2.5? [英] How to round to 2.5 in Java?

问题描述

所以我正在为 android 制作一个健身应用程序，现在我要求用户输入一个数字，例如72.5

So I am making a fitness app for android and now I am asking the user to input a number e.g. 72.5

我会取这个数字并取其百分比并将函数应用于此等.

I would take this number and take percentages of it and apply functions to this etc.

我需要确保我在该数字中所占的百分比四舍五入为 2.5.这是因为在英国健身房你只有以下盘子: 1.25x2=2.5 2.5x2=5 5+2.5=7.5 , 10, 15, 20, 25

I need to make sure that the percentage I take of that number is rounded to 2.5. This is because in a UK gym you only have the following plates: 1.25x2=2.5 2.5x2=5 5+2.5=7.5 , 10, 15, 20, 25

我的意思是它会是这样的数字:40, 42.5, 45, 47.5, 50

What I mean is that it would be numbers like these: 40, 42.5, 45, 47.5, 50

如何将数字 N 舍入到最接近的 2.5?我知道 math.Round() 四舍五入到最接近的整数，但是像这样的自定义数字呢?

How can I round a Number N to the nearest 2.5? I understand that math.Round() rounds to nearest whole but what about to a custom number like this?

推荐答案

请按以下步骤操作:

public class Main {
    public static void main(String args[]) {
        // Tests
        System.out.println(roundToNearest2Point5(12));
        System.out.println(roundToNearest2Point5(14));
        System.out.println(roundToNearest2Point5(13));
        System.out.println(roundToNearest2Point5(11));
    }

    static double roundToNearest2Point5(double n) {
        return Math.round(n * 0.4) / 0.4;
    }
}

输出:

12.5
15.0
12.5
10.0

说明:

下面的例子会更容易理解:

It will be easier to understand with the following example:

double n = 20 / 3.0;
System.out.println(n);
System.out.println(Math.round(n));
System.out.println(Math.round(n * 100.0));
System.out.println(Math.round(n * 100.0) / 100.0);

输出:

6.666666666666667
7
667
6.67

正如您在此处看到的，舍入 20/3.0 返回 7(这是将 0.5 添加到 20 后的下限值/3.0. 检查 this 以了解实现).但是，如果您想将其四舍五入到最近的 1/100 位(即最多 2 个小数位)，一种更简单的方法(但不是那么精确.检查这更精确的方法)将四舍五入n * 100.0(这将使它成为667)然后除以100.0> 这将给出 6.67(即最多 2 个小数位).注意 1/(1/100.0) = 100.0

As you can see here, rounding 20 / 3.0 returns 7 (which is the floor value after adding 0.5 to 20 / 3.0. Check this to understand the implementation). However, if you wanted to round it up to the nearest 1/100th place (i.e. up to 2 decimal places), an easier way (but not so precise. Check this for more precise way) would be to round n * 100.0 (which would make it 667) and then divide it by 100.0 which would give 6.67 (i.e. up to 2 decimal places). Note that 1 / (1 / 100.0) = 100.0

同样，要将数字四舍五入到最近的 2.5 位，您需要对 1/2.5 = 0.4 做同样的事情，即 Math.round(n * 0.4)/0.4.

Similarly, to round the number to the nearest 2.5th place, you will need to do the same thing with 1 / 2.5 = 0.4 i.e. Math.round(n * 0.4) / 0.4.

要将数字四舍五入到最近的 100 位，您需要对 1/100 = 0.01 执行相同的操作，即 Math.round(n * 0.1)/0.1.

To round a number to the nearest 100th place, you will need to do the same thing with 1 / 100 = 0.01 i.e. Math.round(n * 0.1) / 0.1.

要将数字四舍五入到最近的 0.5 位，您需要对 1/0.5 = 2.0 执行相同的操作，即 Math.round(n * 2.0)/2.0.

To round a number to the nearest 0.5th place, you will need to do the same thing with 1 / 0.5 = 2.0 i.e. Math.round(n * 2.0) / 2.0.

我希望很清楚.

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