Accessing Inherited Functions

-

 You are given three classes A, B and C. All three classes implement their own version of func.

In class A, func multiplies the value passed as a parameter by :

class A
{
    public:
        A(){
            callA = 0;
        }
    private:
        int callA;
        void inc(){
            callA++;
        }

    protected:
        void func(int & a)
        {
            a = a * 2;
            inc();
        }
    public:
        int getA(){
            return callA;
        }
};

In class B, func multiplies the value passed as a parameter by :

class B
{
    public:
        B(){
            callB = 0;
        }
    private:
        int callB;
        void inc(){
            callB++;
        }
    protected:
        void func(int & a)
        {
            a = a * 3;
            inc();
        }
    public:
        int getB(){
            return callB;
        }
};

In class C, func multiplies the value passed as a parameter by :

class C
{
    public:
        C(){
            callC = 0;
        }
    private:
        int callC;
        void inc(){
            callC++;
        }
    protected:
        void func(int & a)
        {
            a = a * 5;
            inc();
        }
    public:
        int getC(){
            return callC;
        }
};

You are given a class D:

class D 
{

	int val;
	public:
		//Initially val is 1
		 D()
		 {
		 	val = 1;
		 }


		 //Implement this function
		 void update_val(int new_val)
		 {

			
		 }
		 //For Checking Purpose
		 void check(int); //Do not delete this line.
};

You need to modify the class D and implement the function update_val which sets D's val to new_val by manipulating the value by only calling the func defined in classes A, B and C.

It is guaranteed that new_val has only  and  as its prime factors.

Input Format

Implement class D's function update_val. This function should update D's val only by calling A, B and C's func.

Constraints

 new_val 
Note: The new_val only has  and  as its prime factors.

Sample Input

new_val 

Sample Output

A's func will be called once.
B's func will be called once.
C's func will be called once.

Explanation

Initially, val .

A's func is called once:

val = val*2  
val = 2

B's func is called once:

val = val*3
val = 6

C's func is called once:

val = val*5
val = 30
SOLUTIONS:
#include<iostream>

using namespace std;

class A
{
    public:
        A(){
            callA = 0;
        }
    private:
        int callA;
        void inc(){
            callA++;
        }

    protected:
        void func(int & a)
        {
            a = a * 2;
            inc();
        }
    public:
        int getA(){
            return callA;
        }
};

class B
{
    public:
        B(){
            callB = 0;
        }
    private:
        int callB;
        void inc(){
            callB++;
        }
    protected:
        void func(int & a)
        {
            a = a * 3;
            inc();
        }
    public:
        int getB(){
            return callB;
        }
};

class C
{
    public:
        C(){
            callC = 0;
        }
    private:
        int callC;
        void inc(){
            callC++;
        }
    protected:
        void func(int & a)
        {
            a = a * 5;
            inc();
        }
    public:
        int getC(){
            return callC;
        }
};

class D :public A,B,C
{

    int val;
    public:
        //Initially val is 1
         D()
         {
             val=1;
         }


         //Implement this function
         void update_val(int new_val)
         {
             int a = new_val;
             while(new_val!=0)
             {
                 if(val==a)
                    break;
                 if(new_val%2==0)
                 {
                     A::func(val);
                     new_val/=2;
                 }
                 else if(new_val%3==0)
                 {
                     B::func(val);
                     new_val/=3;
                 }
                 else if(new_val%5==0)
                 {
                     C::func(val);
                     new_val/=5;
                 }
             }

         }
         //For Checking Purpose
         void check(int); //Do not delete this line.
};





void D::check(int new_val)
{
    update_val(new_val);
    cout << "Value = " << val << endl << "A's func called " << getA() << " times " << endl << "B's func called " << getB() << " times" << endl << "C's func called " << getC() << " times" << endl;
}


int main()
{
    D d;
    int new_val;
    cin >> new_val;
    d.check(new_val);

}

Comments

Post a Comment

Popular posts from this blog

Valid Username Regular Expression

-
  You are updating the username policy on your company's internal networking platform. According to the policy, a username is considered valid if all the following constraints are satisfied: The username consists of   to   characters inclusive. If the username consists of less than   or greater than   characters, then it is an invalid username. The username can only contain alphanumeric characters and underscores ( _ ). Alphanumeric characters describe the character set consisting of  lowercase  characters  ,  uppercase  characters  , and digits  . The  first  character of the username must be an  alphabetic  character, i.e., either  lowercase  character   or  uppercase  character  . For example: Username Validity INVALID; Username length < 8 characters VALID VALID INVALID; Username begins with non-alphabetic character INVALID; '?' character not allow...
Read more

Java SHA-256

-
  Cryptographic hash functions are mathematical operations run on digital data; by comparing the computed   hash   (i.e., the output produced by executing a hashing algorithm) to a known and expected hash value, a person can determine the data's integrity. For example, computing the hash of a downloaded file and comparing the result to a previously published hash result can show whether the download has been modified or tampered with. In addition, cryptographic hash functions are extremely collision-resistant; in other words, it should be extremely difficult to produce the same hash output from two different input values using a cryptographic hash function. Secure Hash Algorithm 2  ( SHA-2 ) is a set of cryptographic hash functions designed by the National Security Agency (NSA). It consists of six identical hashing algorithms (i.e.,  SHA-256 ,  SHA-512 ,  SHA-224 ,  SHA-384 ,  SHA-512/224 ,  SHA-512/256 ) with a variable digest size....
Read more

Java Interface

-
  A Java interface can only contain method signatures and fields. The interface can be used to achieve polymorphism. In this problem, you will practice your knowledge on interfaces. You are given an interface  AdvancedArithmetic  which contains a method signature  int divisor_sum(int n) . You need to write a class called MyCalculator which implements the interface. divisorSum  function just takes an integer as input and return the sum of all its divisors. For example divisors of 6 are 1, 2, 3 and 6, so  divisor_sum  should return 12. The value of n will be at most 1000. Read the partially completed code in the editor and complete it. You just need to write the MyCalculator class only.  Your class shouldn't be public. Sample Input 6 Sample Output I implemented: AdvancedArithmetic 12 Explanation Divisors of 6 are 1,2,3 and 6. 1+2+3+6=12. SOLUTION: import  java . util .*; interface  AdvancedArithmetic {    int  divisor_sum ( ...
Read more