题解 | #日期差值#

#include <iostream>
using namespace std;
//计算当年月和日的天数
int GetMonthDay(int year, int month, int day)
{
    int GetMonthday[13] = {0,31,59,90,120,151,181,212,243,273,304,334,365};
    int monthday = GetMonthday[month - 1] + day;
    if((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0) && month > 2)
    {
        return ++monthday;
    }
    return monthday;
}
//判断闰年否，返回当年的天数
int GetYearDay(int year)
{
    int GetYearday[2] = {365,366};
    if((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
    {
        return GetYearday[1];
    }
    return GetYearday[0];
}

int main() {
    int y1,m1,d1,y2,m2,d2;
    int minyear,maxyear;
    int count,n1,n2;
    while(scanf("%04d%02d%02d %04d%02d%02d",&y1,&m1,&d1,&y2,&m2,&d2) != EOF)
    {
        count = 0;
	    //找到小的年，按年相加直到加到与比较年同年
        minyear = y1 < y2 ? y1 : y2;
        maxyear = y1 > y2 ? y1 : y2;
	  	//此时不能等于maxyear,如果等于则多加了一年的天数
        while(minyear < maxyear)
        {
            count += GetYearDay(minyear);
            ++minyear;
        }
        //调用计算当年月和日天数的函数，此时拿到两个同年的月和日的天数：
        n1 = GetMonthDay(y1,m1,d1);
        n2 = GetMonthDay(y2,m2,d2);
        //这一步需要得出Max(n1,n2) - Min(n1,n2)的天数
	  	//最后，把两个日期的年的相差天数+Max(n1,n2) - Min(n1,n2)的天数+1就得出两个日期相差的天数了
        int max = n1 > n2 ? n1 : n2;
        int min = n1 < n2 ? n1 : n2;
        cout << count + (max - min)+1 << endl;
    }
    return 0;    
}