Question

Java question!

Ask the user for the number of boxes to use (N). These boxes will be represented in our program as an array. Each box has only two states – open or closed. So make this an array of Boolean, true = closed, false = open. Start with all boxes closed. Begin with box #2 (remember this is index 1 in the list) and open it and every 2^nd box. Now starting with box 3, and continuing every 3^rd box we will open it if it is closed and close it if it is opened. Go to fourth and reverse the status of every fourth box the same way. Continue opening and closing boxes every 5^th, every 6^th, etc… until we get to where we are starting at box N. Report all the boxes that are closed after we finish our procedure, do not report the open ones. You should see a recognizable pattern with the few that remain closed.

The printed sequence should be: 1 4 9 16 25 36 49……

Answer 1

The executable code for the given program is:

import java.util.Scanner;
class sequence
{
    public static void main(String args[])
    {
        System.out.print("Enter number of boxes:");/*entering no of boxes*/
        int N,i,j,k;
        Scanner sc=new Scanner(System.in);/*creating object for Scanner class for taking input from user*/
        N=sc.nextInt();/*entering no of boxes*/
        boolean arr[]; /*declearing array as boolean to store true(box is closed) and false (box is opened)values */
        arr = new boolean[N];
        for(i=0;i<N;i++)
        {
             arr[i]=true;/*initializing array with all true values.means,all boxes are closed initially*/
        }
        for(i=0;i<N;i++)
        {
            k=i+1;/*we need to start from the second block*/
            for(j=0;j<=N;j++)
            {
                if(j==k && k<N)
                {
                    /*swapping values.if true is present,we will change it as false.if false is present,we will change that as true*/
                    if(arr[k]==true)
                    {
                           arr[k]=false;
                    }
                    else
                    {
                           arr[k]=true;
                    }
                    k=k+i+2;/*this step is the crutial step.if i = 1 we have to increment 2 to the previous value.if i=2 we have to increment 3 to the previous values
                                  and so on.*/
                }
             }
        }
        System.out.println("The Closed boxes are:");
        for(i=0;i<N;i++)
        {
           if(arr[i]==true)
           {
               System.out.println(i+1);/*printing the indices of the boxes that are closed.Here i+1 represents blocks are starting with 1 not 0.*/
           }
        }
    }
}

program alignment in screenshots :

continution.

OUTPUT for the given CODE is: