Efficient algorithm for converting number of days to years (including leap years)

Question

The problem

I am writing a class for holding dates in c++, and I found the following problem:

I have a number of days N since a reference date (in

Answer 1

bool IsLeapYear(int year)
{
    boost::gregorian::date d1(year, 1, 1);
    boost::gregorian::date d2 = d1 + boost::gregorian::years(1);
    boost::gregorian::date_duration diff = d2 - d1;
    return diff.days() != 365;
}