java - do-while 循环出现问题。使用 BlueJ 编译成功

Question

每个完美立方(例如 8、27、216 等)都可以表示为一系列连续奇数的总和。事实上，这样的序列将恰好包含“n”个连续奇数，其中“n”是给定立方体的立方根。

 For example: 1) 8=3+5= 2^3 (2 odd numbers)
              2) 125=21+23+25+27+29= 5^3 (5 odd numbers)
              3) 1000=91+93+95+97+99+101+103+105+107+109= 10^3 (10 odd numbers)

以下代码生成一系列连续奇数，其总和等于作为输入的完美立方(变量名称“cube”)。 问题 - 在给定的代码中没有语法错误，并且尽管存在 do-while 循环，但它仅运行一次，这确保用户可以通过在询问时输入是/否 (Y/N) 来尝试不同的多维数据集。

import java.util.*;

class CUBE {   

    void main() {              
        /*gives a series of consecutive odd numbers                                          
        whose sum is equal to the input value.  
        input is a cube of a number less than 1000*/

        int i,odd,t,sum,cube,n; String c="y"; 
        Scanner sc=new Scanner(System.in);        

        do {
            i=1;
            odd=1;
            t=2;
            sum=0;
            System.out.println("\nEnter the cube");
            cube=sc.nextInt(); //input is stored in 'cube'
            n=cubeRoot(cube);  /*'n' is the cube root of 'cube'. If 'cube'  
                                     is not a perfect cube then n=0*/
            while(i<=n) {
                sum+=odd;  //consecutive odd numbers are are added in sum    
                if(sum==cube) //loop stops if sum=cube
                {
                    break;
                }
                else if (i==n && sum!=cube) {
                    i=1; //counter goes back to 1
                    sum=0;
                    odd=i+t; //odd becomes the next odd number just after 1 and then the one after that
                    t+=2;
                }
                else {
                    i++; 
                    odd+=2;
                }
            }
            if (n!=0) { //if 'cube' is a perfect cube then n will never be 0
                System.out.print("\n"+cube+" = ");
                 for(i=odd-2*(n-1);i<=odd;i+=2)
           {
               if(i==odd)  
                 System.out.print(i+"\n\n");
               else
                 System.out.print(i + " + ");
           }
                System.out.println("\nTry again? (Y/N)\n");                                   
                c=sc.nextLine();
            } 
        }
        while(c.equals("y")||c.equals("Y"));
        //if c is "y" then loop should run again but it doesnt
    }

    int cubeRoot(int cube)  {                    
        /*returns the cube root of 
                    cube and returns 0 if its 
                    invalid */

        int i;
        for(i=1;i<=1000;i++) //n sholud be less than 1000
        {
            if(i*i*i==cube) //if i^3 = cube then n=i
                return i;
        }
        System.out.println("\nINVALID INPUT.");//prints if cube is not a perfect cube 
        return 0;
    }
} 

Answer 1

只需在 while 循环内添加一个新的扫描仪，它就可以工作:

                   System.out.println("\nTry again? (Y/N)\n"); 
                   Scanner sc2=new Scanner(System.in);                                 
                   c=sc2.nextLine();

输入 - 输出:

c is : y

Enter the cube
1

1 = 1

Try again? (Y/N)

Y

Enter the cube
2

INVALID INPUT.

Enter the cube
3

INVALID INPUT.

Enter the cube
4

INVALID INPUT.

Enter the cube
5

INVALID INPUT.

Enter the cube
6

INVALID INPUT.

Enter the cube
7

INVALID INPUT.

Enter the cube
8

8 = 1
2

Try again? (Y/N)

N

正如你所看到的，只要你输入“y”或“Y”它就会继续运行，当你输入时程序退出，如你所愿!

顺便说一句，您的代码无法编译。这是一个编译并成功运行的 Java 类。您的代码无法编译的原因是

1) 你没有正确声明类的main方法2)您没有创建该类的实例，因此您无法按照您尝试的方式调用该方法(这会导致编译错误)。

请参阅以下有效代码:

 import java.util.*;
         class CUBE {   

        public static void main(String[] args)   {              /*gives a series of consecutive odd numbers                                          
                                       whose sum is equal to the input value.  
                                       input is a cube of a number less than 1000*/

                int i,odd,t,sum,cube,n; 
                String c="y"; 
                System.out.println("c is : " + c);
                Scanner sc=new Scanner(System.in);        

                do
                {
                    i=1;
                    odd=1;
                    t=2;
                    sum=0;
                    System.out.println("\nEnter the cube");
                    cube=sc.nextInt(); //input is stored in 'cube'
                    CUBE myCube = new CUBE();
                    n=myCube.cubeRoot(cube);  /*'n' is the cube root of 'cube'. If 'cube'  
                                         is not a perfect cube then n=0*/


                  while(i<=n) 

                    {
                        sum+=odd;  //consecutive odd numbers are are added in sum    
                       if(sum==cube) //loop stops if sum=cube
                        {
                            break;
                        }
                        else if(i==n && sum!=cube)
                        {
                            i=1; //counter goes back to 1
                            sum=0;
                            odd=i+t; //odd becomes the next odd number just after 1 and then the one after that
                            t+=2;
                        }
                        else
                        {
                            i++; 
                            odd+=2;
                        }
                    }
                    if(n!=0) //if 'cube' is a perfect cube then n will never be 0
                    {
                       System.out.print("\n"+cube+" = ");
                       for(i=1;i<=n;i++,odd-=2) //i gives the required odd numbers of the series
                       {
                             System.out.println(i);
                       }
                       System.out.println("\nTry again? (Y/N)\n"); 
                       Scanner sc2=new Scanner(System.in);                                 
                       c=sc2.nextLine();
                    } 
                }
                while(c.equals("y")||c.equals("Y"));//if c is "y" then loop should run again but it doesnt
            }


            int cubeRoot(int cube)  {                    /*returns the cube root of 
                                                          cube and returns 0 if its 
                                                          invalid */

                int i;
                for(i=1;i<=1000;i++)//n sholud be less than 1000
                {
                    if(i*i*i==cube) //if i^3 = cube then n=i
                      return i;
                }
                System.out.println("\nINVALID INPUT.");//prints if cube is not a perfect cube 
                return 0;
            }}