我将如何着手对双方法程序实施 Sentinel 迭代?

How would I go about implementing Sentinel iteration to two-method program?

我创建了一个程序,提示用户输入两个整数并确定第二个整数是否是第一个整数的倍数,并且它适用于一对数字。但是,我想实现哨兵控制的迭代,以便程序重复对用户的查询并产生输出,直到用户想要终止程序。我曾多次尝试使用传统的实现方式来完成此操作,但我 运行 遇到了在重复整个循环时识别和合并退出条件的问题,换句话说,我不确定如何在不破坏程序的情况下将 continue/exit 条件加入程序。

这是程序正常工作的代码,但需要在它能工作的地方实现循环。提前致谢。

import java.util.Scanner; 

public class Multiples {


    public static void main(String[] args) {
        //import scanner
       Scanner input = new Scanner(System.in);
       
     
       System.out.printf ("Enter two integers separated by a space. %n ");
       int number2 = input.nextInt();
       int number1 = input.nextInt();
     
   
       //ask user for input of two integers, to determine if the first integer is - 
       //- a multiple of the second
       
       
       
       //begin method call for ismultiple 
       
       if (ismultiple(number1, number2)==true)
           
           System.out.println (number1+" is a multiple of "+number2);
       else
           System.out.println (number1+" is not a multiple of "+number2);
       
       //end main method
       
       //begin execution of method ismultiple using boolean output for true
    
    }
    public static boolean ismultiple(int x, int y) {
            if(x%y==0)
            return true;
            if (x%y==1)
            return true;
            else
            return false;
    }
      //terminate ismultiple              
    
}
//terminate program
import java.util.Scanner;

class Multiples {

public static void main(String[] args) {
    //import scanner

input();

}

public static void input()
{
  Scanner input = new Scanner(System.in);


  System.out.printf ("Enter two integers separated by a space. %n ");
  int number2 = input.nextInt();
  int number1 = input.nextInt();

  if (ismultiple(number1, number2)==true)

      System.out.println (number1+" is a multiple of "+number2);
  else
      System.out.println (number1+" is not a multiple of "+number2);

//recursion
  System.out.println("want to perform operations again?\n yes-y no-n");
      char choice=input.next().charAt(0);

      switch(choice)
      {
        case 'y': input();
        break;
      }


}
public static boolean ismultiple(int x, int y) {
        if(x%y==0)
        return true;
        if (x%y==1)
        return true;
        else
        return false;
}
  //terminate ismultiple
}

我添加了一个新的输入法并再次调用它,如图所示