题解 | #计算日期到天数转换#

#include <cmath>
#include <iostream>
using namespace std;

bool isleap(int year)
{
    return (year%4==0&&year%100!=0)||year%400==0;
}
//根据年与月返回当月的天数
int GetDaysByYM(int year,int month)
{
    int days[13]={29,31,28,31,30,31,30,31,31,30,31,30,31};
    if(month==2&&isleap(year))
    {
        month=0;
    }
    return days[month];
}
//根据年月日返回当前日期是当年的第几天
int GetdateByYMD(int year,int month,int day)
{
    int days=0;
  //循环加上月数减一的整天数
    for(int i=1;i<month;++i){
        days+=GetDaysByYM(year,i);
    }
  //最后加上当月已经过的天数day
    days+=day;
    return days;
}

int main() {
    int year,month,day;
    while (cin >> year >> month>>day) {
        day=GetdateByYMD(year,month,day);
        cout<<day<<endl;
    }
}