题解 | 日期累加

#include <iostream>
using namespace std;
class date
{
public:
    friend ostream& operator<<(ostream& out, const date& d);
    date operator+(int n);
    date(int year,int month,int day);
    int GetMonthDay(int year, int month);
private:
    int _year;
    int _month;
    int _day;
};

ostream& operator<<(ostream& out,const date& d)
{
    printf("%04d-%02d-%02d",d._year,d._month,d._day);
    return out;
}

int date::GetMonthDay(int year,int month)
{
    static int monthday[13]={-1,31,28,31,30,31,30,31,31,30,31,30,31};
    if(month==2&&
    ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0))
    {
        return 29;
    }
    return monthday[month];
}
date::date(int year,int month,int day)
{
    _day=day;
    _month=month;
    _year=year;
}
date date::operator+(int n)
{
    int year=this->_year;
    int month=this->_month;
    int day=this->_day;

    while(day+n>GetMonthDay(year, month))
    {
        n-=GetMonthDay(year, month);
        month++;
        if(month>12)
        {
            year++;
            month=1;
        }
    }
    day+=n;
    return date(year,month,day);
}
int main() 
{
    int m,year,month,day,n;
    cin>>m;
    while(m--)
    {
        while(cin>>year>>month>>day>>n)
        {
            date d(year,month,day);
            date d1=d+n;
            cout<<d1<<endl;
        }
    }

}