Algorithm
Problem Name:
In this HackerRank Functions in C++ programming problem solution,
You are given a class - Complex.
class Complex
{
public:
int a,b;
};
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 need to overload operators + and << for the Complex class.
The operator + should add complex numbers according to the rules of complex addition:
(a+ib)+(c+id) = (a+c) + i(b+d)
Overload the stream insertion operator << to add "
" to the stream:
cout<<c<<endl;
The above statement should print "a+ib" followed by a newline where a =c*a and b = c*b.
Input Format
The overloaded operator + should receive two complex numbers (a + ib and c + id) as parameters. It must return a single complex number.
The overloaded operator << should add "a+ib" to the stream where a is the real part and b is the imaginary part of the complex number which is then passed as a parameter to the overloaded operator.
Output Format
As per the problem statement, for the output, print "a+ib" ollowed by a newline where a =c*a and b = c*b.
Sample Input
3+i4
5+i6
Sample Output
8+i10
Explanation
Given output after performing required operations (overloading + operator) is 8+i10.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
//Operator Overloading
#include
using namespace std;
class Complex
{
public:
int a,b;
void input(string s)
{
int v1=0;
int i=0;
while(s[i]!='+')
{
v1 = v1*10 + s[i]-'0';
i++;
}
while(s[i] ==' ' || s[i]=='+'||s[i] == 'i')
{
i++;
}
int v2 = 0;
while(i < s.length())
{
v2 = v2 * 10 + s[i] - '0';
i++;
}
a=v1;
b=v2;
}
};
Complex operator+(const Complex A, const Complex B) {
Complex Z;
Z.a = A.a + B.a;
Z.b = A.b + B.b;
return Z;
}
// (a+ib) + (c+id) = (a+c) + i(b+d)
// a+ib
ostream& operator << (ostream& os, const Complex C) {
return os << C.a << "+" << "i" << C.b;
}
int main()
{
Complex x,y;
string s1,s2;
cin >> s1;
cin >> s2;
x.input(s1);
y.input(s2);
Complex z = x + y;
cout << z << endl;
}
Input
Output