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maw [93]
4 years ago
15

Write a copy constructor for carcounter that assigns origcarcounter.carcount to the constructed object's carcount. sample output

for the given program:
Computers and Technology
2 answers:
Drupady [299]4 years ago
8 0

#include <iostream>

using namespace std;

class CarCounter {

  public:

     CarCounter();

     CarCounter(const CarCounter& origCarCounter);

     void SetCarCount(const int count) {

         carCount = count;

     }

     int GetCarCount() const {

         return carCount;

     }

  private:

     int carCount;

};

CarCounter::CarCounter() {

  carCount = 0;

  return;

}

CarCounter::CarCounter(const CarCounter &p){

carCount = p.carCount;

}

void CountPrinter(CarCounter carCntr) {

  cout << "Cars counted: " << carCntr.GetCarCount();

  return;

}

int main() {

  CarCounter parkingLot;

  parkingLot.SetCarCount(5);

  CountPrinter(parkingLot);

  return 0;

}

Sample output:  

Cars Counted: 5

katen-ka-za [31]4 years ago
7 0

Answer:

import java.util.Comparator;

import java.util.PriorityQueue;

 

public class PriorityQueueTest {

 

static class PQsort implements Comparator<Integer> {

 

 public int compare(Integer one, Integer two) {

 return two - one;

 }

}

 

public static void main(String[] args) {

 int[] ia = { 1, 10, 5, 3, 4, 7, 6, 9, 8 };

 PriorityQueue<Integer> pq1 = new PriorityQueue<Integer>();

 

 // use offer() method to add elements to the PriorityQueue pq1

 for (int x : ia) {

 pq1.offer(x);

 }

 

 System.out.println("pq1: " + pq1);

 

 PQsort pqs = new PQsort();

 PriorityQueue<Integer> pq2 = new PriorityQueue<Integer>(10, pqs);

 // In this particular case, we can simply use Collections.reverseOrder()

 // instead of self-defined comparator

 for (int x : ia) {

 pq2.offer(x);

 }

 

 System.out.println("pq2: " + pq2);

 

 // print size

 System.out.println("size: " + pq2.size());

 // return highest priority element in the queue without removing it

 System.out.println("peek: " + pq2.peek());

 // print size

 System.out.println("size: " + pq2.size());

 // return highest priority element and removes it from the queue

 System.out.println("poll: " + pq2.poll());

 // print size

 System.out.println("size: " + pq2.size());

 

 System.out.print("pq2: " + pq2);

 

}

}

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Given the following adjacency matrix, what is the approximate rank vector after one iteration of the power iteration method (use
ValentinkaMS [17]

Note: The matrix referred to in the question is: M = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]

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a) [5/18, 5/18, 4/9]'

Explanation:

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To start the power iteration, let us start with an initial non zero approximation,

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