Project Euler 39 - Integer right triangles

/*
If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
{20,48,52}, {24,45,51}, {30,40,50}
For which value of p ≤ 1000, is the number of solutions maximised?
*/


#include <iostream>
#include <stdlib.h>
#include <math.h>
#include <vector>
#include <algorithm>
#define LIMITPERIMETER 1000

using namespace std;

int main()
{
    // 1. First, I will create a vector, in which i will place the counter of solution of each perimeter
    // In index = 0 there will be the number of solutions when perimeter = 3 = min, in index = 1, the solutions of perimeter = 4, ....
    // Therefore, the length will be: length = LIMITPERIMETER  - 3 + 1
    vector <int> counterOfPerimSolutions = vector <int> ();
    // Initialization of the vector
    for(int i=0; i<LIMITPERIMETER-3+1; i++){
        counterOfPerimSolutions.push_back(0);
    }

    // 2. I make 2 for-loops
    for(int a=1; a<=999; a++){
        for(int b=1; b<=999; b++){
            double c = sqrt(a*a + b*b);
            if(c == int(c) && a+b+c <= LIMITPERIMETER){
                    // That means that a,b,c are integers, that create
                    int perimeter =  a + b + c;
                    // This value is in index perimeters - 3 in the vector
                    cout << "a = " << a << ", b = " << b << ", c =  " << c << " ----> perimeter = " << perimeter << endl;
                    counterOfPerimSolutions[perimeter-3]++;
            }
        }
    }

    /*
    for(int i=0; i<counterOfPerimSolutions.size(); i++){
        cout << counterOfPerimSolutions[i] << " sols for perim = " << i+3 << endl;
    }
    */

    // 3. Now, the vector contains all the solutions, but each solution is twice, because there are 2 different combinations of a and b (and therefore c), that lead to the same perimeter
    // For perimeter = 120, I can have a=30, b=40, c=50 and a=40, b=30, c=50, but that's the same solution
    for(unsigned int i=0; i<counterOfPerimSolutions.size(); i++){
        counterOfPerimSolutions[i] /= 2;
    }

    // 4. Find the perimeter, with the most solutions
    int maxIndex = 0;
    int maxSolutions = counterOfPerimSolutions[0];
    for(unsigned int i=1; i<counterOfPerimSolutions.size(); i++){
        if(counterOfPerimSolutions[i] > maxSolutions){
            maxSolutions = counterOfPerimSolutions[i];
            maxIndex = i;
        }
    }

    // 5. Now, I have the value of maxIndex. But this index is connected with a perimeter's value, especially the value: maxIndex + 3
    cout << "\nIn range: 3 <= perimeter <= " << LIMITPERIMETER << ", the max different solutions are " << counterOfPerimSolutions[maxIndex] << " and the value of this perimeter is " << maxIndex + 3 << endl;
    return 0;

}