1.题目：

 

2.分析：

1. 分别求出每一个日期与0000年0月1日距离的天数

2. 两个距离天数相减即可得到两个日期相差的天数

3.我的代码：

#include <iostream>
using namespace std;

class Date 
{
public:
    Date(int year = 1, int month = 1, int day = 1)
    : _year(year)
    , _month(month)
    , _day(day) {}


    // 获取天数
    inline int GetMonthDay(int year, int month) 
    {
        static int arr[13] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
        if ((month == 2) && ((year % 400 == 0) || ((year % 4 == 0) && (year % 100 != 0))))
        {
            return 29;
        }
        else 
        {
            return arr[month];
        }
    }

    bool IsSame(Date& date) 
    {
        return this->_year == date._year
            && this->_month == date._month
            && this->_day == date._day;
    }

    void Gap(int year, int month, int day)
    {
        int n = 1;
        Date date(year, month, day);
        while (IsSame(date) == false)
        {
            n++;
             _day++;
            while (_day > GetMonthDay(_year, _month))
            {
                _day -= GetMonthDay(_year, _month);
                ++_month;
                if (_month > 12)
                {
                    _month -= 12;
                    ++_year;
                }
            }
        }
        cout << n << endl;
    }
private:
    int _year;
    int _month;
    int _day;
};
int main() 
{
    int year1 = 0;
    int month1 = 0;
    int day1 = 0;
    int year2 = 0;
    int month2 = 0;
    int day2 = 0;
    char arr1[9];
    char arr2[9];
    cin >> arr1 >> arr2;
    sscanf(arr1,"%4d%2d%2d",&year1,&month1,&day1);//格式化读入
    sscanf(arr2,"%4d%2d%2d",&year2,&month2,&day2);
    Date date(year1, month1, day1);
    date.Gap(year2, month2, day2);
}

4.答案代码：

 #include <iostream>

  using namespace std;

  /*

  *思路：

  * 1. 分别求出每一个日期与0000年0月1日距离的天数

  * 2. 两个距离天数相减即可得到两个日期相差的天数

  */

  //平年从1月到n月的天数

  int mon[12]={0,31,59,90,120,151,181,212,243,273,304,334};

  //给出年月日，计算距离0000年0月1日的天数和

  

  int CountDay(int y,int m,int d)

  {

    // 计算0-y年的天数

    int yearDay = y*365+y/4-y/100+y/400;

    // 计算到0-m月的天数

    int monthDay = mon[m-1];

    

    if(m > 2 && ((y%4 == 0 && y%100 != 0) || y%400==0))

      monthDay += 1;

    

    return yearDay + monthDay + d;

  }

  

  int main()

  {

    int year1,month1,day1;

    scanf("%4d%2d%2d",&year1,&month1,&day1);

    int n1 = CountDay(year1,month1,day1);

    

    int year2,month2,day2;

  scanf("%4d%2d%2d",&year2,&month2,&day2);

    int n2 = CountDay(year2,month2,day2);

    

    cout<<abs(n1-n2)+1<<endl;

  }