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guapka [62]
3 years ago
9

SHOW ALL YOUR WORK. REMEMBER THAT PROGRAM SEGMENTS ARE TO BE WRITTEN IN JAVA. Assume that the classes listed in the Java Quick R

eference have been imported where appropriate. Unless otherwise noted in the question, assume that parameters in method calls are not null and that methods are called only when their preconditions are satisfied. In writing solutions for each question, you may use any of the accessible methods that are listed in classes defined in that question. Writing significant amounts of code that can be replaced by a call to one of these methods will not receive full credit. This question involves a game that is played with multiple spinners. You will write two methods in the SpinnerGame class below. public class SpinnerGame { /** Precondition: min < max * Simulates a spin of a spinner by returning a random integer * between min and max, inclusive. */ public int spin(int min, int max) { /* to be implemented in part (a) */ } /** Simulates one round of the game as described in part (b). */ public void playRound() { /* to be implemented in part (b) */ } } (a) The spin method simulates a spin of a fair spinner. The method returns a random integer between min and max, inclusive. Complete the spin method below by assigning this random integer to result. /** Precondition: min < max * Simulates a spin of a spinner by returning a random integer * between min and max, inclusive. */ public int spin(int min, int max) { int result; return result; }

Computers and Technology
1 answer:
adoni [48]3 years ago
4 0

Answer:

Check the explanation

Explanation:

Player 1 Coin

Player 2 Coin

Player 1

Player 2

Round

Count at

Count at

Player

Player

Coin

Coin

Number

Beginning of

Beginning of

1

2

Outcome

Count at

Count at

Round

Round

Spends

Spends

End of

End of

Round

Round

Off-by-

one,

10 - 1

10

10

2

player 2

10 - 2

+

= 8

1

gains 1

= 10

coin

Same,

10 - 2

2

8

10

2

2

player 2

-

gains 1

6

+

=

1

=

9

coin

Off-by-

two,

6 - 1

+

3

6

9

3

player 1

2

9 - 3

gains 2

=

=

7

6

coins

Same,

4

7

6

2

2

player 2

7 - 2

6 - 2+

gains 1

=

5

=

coin

5

Kindly check the attached image below to see the well arranged table to solve the above question.

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Artists who draw images on a computer often use a stylus in conjunction with special _________ tablets, which offer extra-sensit
hodyreva [135]

Answer:

Graphics is the correct answer to the following question.

Explanation:

The following answer is true because Graphic tablet is the device in which a person can draw graphics, images or they can also draw by their hands with use of digital pen which is use for drawing as they draw images on the paper. we can also say the graphic tablet is used for the digital art.

So, That's why the following answer is correct.

6 0
3 years ago
Create an instance of your queue that accepts java.lang.String Objects. Starting with an empty queue, use the enqueue(E e) metho
Aleksandr-060686 [28]

Answer:

Code is given below:

Explanation:

ode.java:

public class Node

{

String data;

Node next;

Node prev;

public Node(String data,Node next, Node prev){

this.next=next;

this.data=data;

this.prev=prev;

}

public Node(){

}

public String getData(){

return data;

}

public void setData(String data){

this.data=data;

}

public Node getNext(){

return next;

}

public void setNext(Node next){

this.next=next;

}

public Node getPrev(){

return prev;

}

public void setPrev(Node prev){

this.prev=prev;

}

}

Linked Queue.java:

public class LinkedQueues {

    Node head ;

   Node tail;

int size=0;

public Oueues(){

   this.head=null;

   this.tail=null;

}

public boolean isEmpty()

{

   return head == tail;

}    

 public int getSize()

    {

          return size;

  }    

 public void insert(String data){

   Node tmp = new Node(data,null,null);

    tmp.data=data;

    tmp.next=null;

    if(head==null){

        head=tail=tmp;

        head.prev=null;

    }

    else{

        tail.next=tmp;

        tmp.prev=tail;

        tail=tmp;

 }    

 }

 public String remove(){

  if(head.next==tail)

      return null;// list empty

  Node tmp=head.next;

  head.next=tmp.next;

  tmp.next.prev=head;

  list();

  return tmp.data;

 }

 public void list(){

     System.out.println("Queues");

     if(size==0){

         System.out.println("İs Empty");

     }

    Node tmp=head;

    while(tmp !=tail.getNext()){

        System.out.println(tmp.getVeri()+" ");

      tmp= tmp.getNext();

    }

     System.out.println();

 }

Linkedstack.java:

public class LinkedStack {

Node head = null;

Node tail = null;

int size=0;

     public int getSize() {

    return size;

      }

  public boolean isEmpty()

      {

   return head == null;

   }    

 public void Push(String data) {

    tail = head;

   head = new Node(data,null,null);

   head.data=data;

   head.next= tail;

   head.prev = null;

   if(tail != null) {

       tail.prev=head;

   }

   size++;

 }

 public void Pop() {

   if (!isEmpty()) {

       head = head.next;   // delete first node

       size--;

   } else {

       System.out.println("İs Empty");

   }

}

  public void Top() {

   Node tmp = head;

   while (tmp != null) {

       System.out.println(tmp.getData());

       tmp = tmp.getNext();

   }

}

 }

ArrayBasedQueue.java

import java.util.LinkedList;

import java.util.Queue;

 

public class ArrayBasedQueue

{

 public static void main(String[] args)

 {

   Queue<Integer> q = new LinkedList<>();

 

   // Adds elements {0, 1, 2, 3, 4} to queue

   for (int i=0; i<5; i++)

    q.add(i);

 

   // Display contents of the queue.

   System.out.println("Elements of queue-"+q);

   int removedele = q.remove();

   System.out.println("removed element-" + removedele);

 

   System.out.println(q);

   int head = q.peek();

   System.out.println("head of queue-" + head);

 

   int size = q.size();

   System.out.println("Size of queue-" + size);

 }

}

ArrayBasedStack.java:

import java.io.*;

import java.util.*;

 

public class ArrayBasedStack

{

static void stack_push(Stack<Integer> stack)

{

for(int i = 0; i < 5; i++)

{

stack.push(i);

}

}

 

// Popping element from the top of the stack

static void stack_pop(Stack<Integer> stack)

{

System.out.println("Pop :");

 

for(int i = 0; i < 5; i++)

{

Integer y = (Integer) stack.pop();

System.out.println(y);

}

}

 }

TestTimes.java:

public class TestTimes implements TestTimesInterface

{

 public static enum TimeUnits {};

 public static enum MemoryUnits{};

}

Driver.java:

public class driver

{

public static void main(String args[])

{

   Scanner s = new Scanner(System.in);

   LinkedStack y = new LinkedStack();

   LinkedQueues k = new LinkedQueues();

   ArrayBasedStack as=new ArrayBasedStack();

     ArrayBasedQueue as=new ArrayBasedQueue();

   FileWriter fwy;

   FileWriter fwk;

   File stack = new File("stack.txt");

   if (!stack.exists()) {

       stack.createNewFile();

   } else {

       System.out.println("already exists ");

   }

   BufferedReader reader = null;

   reader = new BufferedReader(new FileReader(stack));

   String line = reader.readLine();

   while (line != null) {

       y.Push(line = reader.readLine());

       System.out.println(line);

   }

   File queue = new File("queue.txt");

   if (!queue.exists()) {

       queue.createNewFile();

   } else {

       System.out.println("already exists ");

   }

   BufferedReader read = null;

   read = new BufferedReader(new FileReader(queue));

   String lines = read.readLine();

   while (lines != null) {

       lines = read.readLine();

       k.insert(lines);

       System.out.println(lines);

   }

int choice;

     System.out.println("1. Stack out- queue add");

     System.out.println("2. Stack add- queue out");

     System.out.println("3. Stack and queue ");

     System.out.println("4. File writer");

     choice = s.nextInt();

   switch (choice) {

       case 1:

           k.insert(s.next());

           k.list();

           y.pop();

           break;

       case 2:

         y.Push(s.next());

           y.Top();

         k.remove();

           break;

       case 3:

           y.Top();

           k.list();

           break;

       case 4:

           fwy = new FileWriter(stack);

           Node no = y.head;

           while (no.next != null) {

               fwy.write("\n" + no.data);

               no = no.next;

           }

           fwy.flush();

           fwy.close();

           fwk = new FileWriter(queue);

           Node noo = k.head;

           while (noo.next != null) {

               fwk.write("\n" + noo.data);

               noo = noo.next;

           }

           fwk.flush();

           fwk.close();

           break;

        }

 

}

}

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