英文:
How to round to 2.5 in Java?
问题
所以我正在制作一个适用于安卓的健身应用，现在我要求用户输入一个数字，例如72.5。
我会取这个数字的百分比，并对其应用函数等等。
我需要确保我取得的这个数字的百分比是以2.5为单位四舍五入的。这是因为在英国的健身房里，你只有以下这些重量片：1.25x2=2.5，2.5x2=5，5+2.5=7.5，10，15，20，25。
我的意思是，会有这些数字：40，42.5，45，47.5，50。
如何将一个数字N四舍五入到最近的2.5？我知道math.Round()会四舍五入到最近的整数，但是对于这样的自定义数字该怎么办呢？
英文:
So I am making a fitness app for android and now I am asking the user to input a number e.g. 72.5
I would take this number and take percentages of it and apply functions to this etc.
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
What I mean is that it would be numbers like these: 40, 42.5, 45, 47.5, 50
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?
答案1
得分: 1
按照以下方式执行：
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
解释：
通过以下示例更容易理解：
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，四舍五入 20 / 3.0 返回 7（这是在将 0.5 加到 20 / 3.0 后的底值。请查看此处了解实现方式）。然而，如果您想将其四舍五入到最接近的 1/100 位置（即到小数点后 2 位），一种更简单的方法（但不太精确。查看此处以获取更精确的方法）是将 n * 100.0 四舍五入（这将使其变为 667），然后除以 100.0，这将得到 6.67（即到小数点后 2 位）。请注意，1 / (1 / 100.0) = 100.0
同样，要将数字四舍五入到最接近的 2.5 位置，您需要使用 1 / 2.5 = 0.4，即 Math.round(n * 0.4) / 0.4。
要将数字四舍五入到最接近的 100 位置，您需要使用 1 / 100 = 0.01，即 Math.round(n * 0.1) / 0.1。
要将数字四舍五入到最接近的 0.5 位置，您需要使用 1 / 0.5 = 2.0，即 Math.round(n * 2.0) / 2.0。
我希望这样清楚了解了这个过程。
英文:
Do it as follows:
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;
}
}
Output:
12.5
15.0
12.5
10.0
Explanation:
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);
Output:
6.666666666666667
7
667
6.67
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
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.
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.
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.
I hope, it is clear.
答案2
得分: 1
晚了回复，但你也可以这样做：2.5 *（Math.round(number/2.5)）
同样的方法，如果你想要将磅数四舍五入到最近的5，可以用这个公式：5 *（Math.round(number/5)）
英文:
Late reply but you can also do 2.5*(Math.round(number/2.5))
Same way if you wanted to round in lbs to the nearest 5th its 5*(Math.round(number/5))
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