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

This

bool IsLeapYear(int year) 
{ 
    if ((year % 4 == 0) && (year % 100 != 0) && (year % 400 == 0)) return true; 
    else return false; 
}

is incorrect. It returns false for 2000. Better:

bool IsLeapYear(int year) 
{ 
    if (year % 400 == 0) return true; 
    if ((year % 4 == 0) && (year % 100 != 0)) return true; 
    return false; 
}