做循环的麻烦。 使用BlueJ成功编译

Question

Every perfect cube (for eg. 8, 27, 216 etc) can be expressed as the sum of a series of consecutive odd numbers. It is a fact that such a series will contain exactly 'n' consecutive odd numbers where 'n' is the cube root of the given cube.

 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)

The following code generates a series of consecutive odd numbers whose sum is equal to a perfect cube which is taken as input (variable name 'cube'). PROBLEM-In the given code there are NO syntax errors and it runs only once inspite of a do-while loop which ensures that the user can try different cubes by entering Yes/No (Y/N) when asked.

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

just add a new scanner inside the while loop, and it works:

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

INPUT - OUTPUT :

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

as you can see, as long as you input "y" or "Y" it keeps on running, when you input the program exits, as you wanted to do!

By the way, your code does not compile. Here is a Java class which compiles and runs successfully. The reason your code could not be compiled, was that

1) you did not declare the main method of the class correctly 2) you did not create an instance of the class, so you could not call the method the way you were trying to (it led to compilation errors).

See the following code which works:

 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;
            }}