Question

我创建了一个Java程序，该程序接受三个整数，并将打印出这三个整数的和，乘积，最小和最大以及平均值。

import java.util.Scanner;

public class ThreeNums
{
    public static void main (String[] args)
    {
        new ThreeNums();
    }

    public ThreeNums ()
    {
        int num1, num2, num3, sum, smallest, largest, exitPrompt = 0;
        float average;

        Scanner scan = new Scanner (System.in);

        while (exitPrompt == 0)
        {
            System.out.printf ("\n\nPlease enter three integers.\n");
            System.out.printf ("\t1st integer: ");
            num1 = scan.nextInt();
            System.out.printf ("\t2nd integer: ");
            num2 = scan.nextInt();
            System.out.printf ("\t3rd integer: ");
            num3 = scan.nextInt();
            System.out.printf ("\n");

            sum = num1+num2+num3;
            System.out.println ("Sum: " + sum);
            System.out.printf ("Product: %d\n", num1*num2*num3);

            if (num1<num2)
                smallest = num1;
            else
                smallest = num2;

            if (smallest<num3)
                smallest = smallest;
            else
                smallest = num3;

            if (num1>num2)
                largest = num1;
            else
                largest = num2;

            if (largest>num3)
                largest = largest;
            else
                largest = num3;

            System.out.printf ("Smallest integer: %d\n", smallest);
            System.out.printf ("Largest integer: %d\n", largest);

            average = sum/3;
            System.out.printf ("Average: %.3f\n", average);

            System.out.printf ("\nExit? 0 if NO, any other digit if YES\n");
            exitPrompt = scan.nextInt();
        }

        System.out.printf ("\n\n");
    }
}

除了（我认为）第56-57行之外，一切都很好：

average = sum/3;
System.out.printf ("Average: %.3f\n", average);

应该打印用户将输入的三个整数的平均值。例如，如果用户输入5、7和7，则应该显示平均值为6.333：

Please enter three integers.
1st integer: 5
2nd integer: 7
3rd integer: 7

Sum: 19
Product: 245
Smallest integer: 5
Largest integer: 7
Average: 6.333

Exit? 0 if NO, any other digit if YES

但是，平均而言，它仅显示6.000。

.
.
.
Average: 6.000
.
.

如果我输入78、45和23，则仅显示48.000，而不是48.667，依此类推。我的代码可能是什么问题？请帮忙...谢谢！

Answer 1

此处：df %>% mutate(ID_rleid = with(rle(type), rep(seq_along(lengths), lengths)))和average = sum/3;均为sum的{{1}}，因此结果为整数，然后将其赋予3。您可以通过以下方法克服这一点：

int

我的平均公式未给出确切答案

1 个答案: