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say hello world with C++ - Solution in Hacker Rank - hackerranksolutions8

  Objective This is a simple challenge to help you practice printing to  stdout . You may also want to complete  Solve Me First  in C++ before attempting this challenge. We’re starting out by printing the most famous computing phrase of all time! In the editor below, use either  printf  or  cout  to print the string  Hello ,World!  to  stdout . The more popular command form is  cout . It has the following basic form: cout<<value_to_print<<value_to_print; Any number of values can be printed using one command as shown. The  printf  command comes from C language. It accepts an optional format specification and a list of variables. Two examples for printing a string are: printf("%s", string);   printf(string); Note that neither method adds a newline. It only prints what you tell it to. Output Format Print   Hello ,World!   to stdout. Sample Output Hello, World! Solution:- //Say Hello, ...

Operator Overloading in C++ – Solution in Hacker Rank - hackerranksolutions8

 

Problem

Classes define new types in C++. Types in C++ not only interact by means of constructions and assignments but also via operators. For example:

int a=2, b=1, c;
c = b + a;

The result of variable c will be 3.
Similarly, classes can also perform operations using operator overloading. Operators are overloaded by means of operator functions, which are regular functions with special names. Their name begins with the operator keyword followed by the operator sign that is overloaded. The syntax is:

type operator sign (parameters) { /*... body ...*/ }


You are given a main() function which takes a set of inputs to create two matrices and prints the result of their addition. You need to write the class Matrix which has a member a of type vector<vector<int> >. You also need to write a member function to overload the operator +. The function’s job will be to add two objects of Matrix type and return the resultant Matrix.


Input Format :

First line will contain the number of test cases T. For each test case, there are three lines of input. The first line of each test case will contain two integers N and M which denote the number of the rows and columns respectively of the two matrices that will follow on the next two lines. These next two lines will each contain N*M elements describing the two matrices in row-wise format i.e. first M elements belong to the first row,next M elements belong to the second row and so on.

Constraints :

  • 1 <= T <= 1000
  • 1 <= N <= 100
  • 1 <= M <= 100
  • 1 <= Aij <= 10, where Aij is the element in the ith row and jth Column of the Matrix.

Output Format :

The code provided in the editor will use your class Matrix and overloaded operator function to add the two matrices and give the output.


Sample Input :

1
2 2
2 2 2 2
1 2 3 4

Sample Output :

3 4 
5 6

Explanation :

The sum of first matrix and the second matrix is the matrix given in the output.


Solution :

//Operator Overloading in C++ - Hacker Rank Solution
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

/* Operator Overloading in C++ - Hacker Rank Solution START */

class Matrix 
{
    public:
        vector<vector<int>> a;

        Matrix() {}
        Matrix operator+(const Matrix &o) 
        {
            Matrix sum(*this);
            for (size_t i = 0; i < sum.a.size(); i++) 
            {
                for (size_t j = 0; j < sum.a[i].size(); j++) 
                {
                    sum.a[i][j] += o.a[i][j];
                }
            }
            return sum;
        }
};

/* Operator Overloading in C++ - Hacker Rank Solution END */

int main () 
{
   int cases,k;
   cin >> cases;
   for(k=0;k<cases;k++) 
   {
      Matrix x;
      Matrix y;
      Matrix result;
      int n,m,i,j;
      cin >> n >> m;
      for(i=0;i<n;i++) 
      {
         vector<int> b;
         int num;
         for(j=0;j<m;j++) 
         {
            cin >> num;
            b.push_back(num);
         }
         x.a.push_back(b);
      }
      for(i=0;i<n;i++) 
      {
         vector<int> b;
         int num;
         for(j=0;j<m;j++) 
         {
            cin >> num;
            b.push_back(num);
         }
         y.a.push_back(b);
      }
      result = x+y;
      for(i=0;i<n;i++) 
      {
         for(j=0;j<m;j++) 
         {
            cout << result.a[i][j] << " ";
         }
         cout << endl;
      }
   }  
   return 0;
}

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