連続シーケンスを探してnに等しくする

#include <iostream>
using namespace std;

void FindContinuousSum(int n)
{
    if (n < 3)
        return;
    int val = (n + 1) / 2;
    int l = 1;
    int h = 2;
    int sum = l + h;
    while (l < val)
    {
        if (sum < n)
        {
            ++h;
            sum += h;
        }
        else if (sum > n)
        {
            sum -= l;
            ++l;
        }
        else
        {
            for (int i = l; i <= h; ++i)
                cout << i << " ";
            cout << endl;
            ++h;
            sum += h;
            sum -= (l + l + 1);
            l += 2;
        }
    }
}

int main()
{
    FindContinuousSum(99);
    return 0;
}