HackerRank Overload Operators solution in C++ programming

In this HackerRank Overload Operators problem solution in the C++ programming language You need to overload operators + and << for the Complex class.

HackerRank Overload Operators problem solution in c++ programming.

//Operator Overloading

#include<iostream>

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;
    }
};

//Overload operators + and << for the class complex
//+ should add two complex numbers as (a+ib) + (c+id) = (a+c) + i(b+d)
//<< should print a complex number in the format "a+ib"
ostream& operator<<(ostream& os, const Complex& c) {
    return os << c.a << (c.b > 0 ? '+' : '-') << 'i' << c.b;
}

Complex operator+(const Complex& a, const Complex& b) { 
    return { a.a + b.a, a.b + b.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;
}

Second solution

//Operator Overloading

#include<iostream>

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;
    }
};


//<< should print a complex number in the format "a+ib"

//Overload operators + and << for the class complex
//+ should add two complex numbers as (a+ib) + (c+id) = (a+c) + i(b+d)
Complex operator +(const Complex &x, const Complex &y) {
    Complex z;
    z.a = x.a + y.a; z.b = x.b + y.b;
    return z;
}
//<< should print a complex number in the format "a+ib"
std::ostream &operator <<(std::ostream& os, const Complex &z) {
    os << z.a << "+i" << z.b;
    return os;
}

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;
}