Question

如果扫描仪输入为负，则不显示任何内容。如果我输入-11那么10，-10和-1应该是输出。

import java.util.Scanner;
public class Factor
{
    public static void main(String[] args)
    {
        Scanner scan = new Scanner(System.in);
        String replay = "replay";
        while(replay.equals("replay"))
        {
        System.out.println("The numbers that add to be ___(number1)___ that multiply to be ___(number2)___ are...");
        System.out.println("Enter number1:");
        int n1 = scan.nextInt();
        System.out.println("Enter number2:");
        int n2 = scan.nextInt();

        System.out.println("Computing...");
        for(double f2 = -1000; f2 <= n1; f2++)
        {
            for(double f1 = -1000; f1 <= n1; f1++)
            {
                if(f1*f2 == n2 && f1+f2 == n1)
                {
                    System.out.println(f1 + " and " + f2);
                }
            }
        }    
        scan.nextLine();
        System.out.println("Enter replay if you would like to compute again");
        replay = scan.nextLine();
        }

    }
}

即使我的循环变量是负数。

Answer 1

您的扫描仪可以使用负数，它运行正常。你没有得到预期的输出，因为你的循环结束于n1是-11，所以循环不会达到（f1 * f2 == n2＆amp;＆amp; f1 + f2 == n1）为真的点。如果你迭代，让我们说从-1000到1000，你将获得所需的输出。

此代码：

`import java.util.Scanner;

public class NegativeScanner{
    public static void main(String[] args)
    {
        Scanner scan = new Scanner(System.in);
        String replay = "replay";
        while(replay.equals("replay"))
        {
        System.out.println("The numbers that add to be ___(number1)___ that multiply to be ___(number2)___ are...");
        System.out.println("Enter number1:");
        int n1 = scan.nextInt(); //-11
        System.out.println("Enter number2:");
        int n2 = scan.nextInt(); // 10

        System.out.println("Computing...");
        for(double f2 = -1000; f2 <= 1000; f2++){ //-1000-től -11-ig

            for(double f1 = -1000; f1 <= 1000; f1++){ //-1000-től -11-ig

                if(f1*f2 == n2 && f1+f2 == n1)
                {
                    System.out.println(f1 + " and " + f2);
                }
            }
        }    
        scan.nextLine();
        System.out.println("Enter replay if you would like to compute again");
        replay = scan.nextLine();
        }
    }
}

产生此输出：

The numbers that add to be ___(number1)___ that multiply to be ___(number2)___ are... Enter number1: -11 Enter number2: 10 Computing... -1.0 and -10.0 -10.0 and -1.0 Enter replay if you would like to compute again

Answer 2

f1和f2的值小于-11。也许对于循环的边界会更好java.lang.Math.max（n1，n2）

为什么负输入不适用于for循环

2 个答案: