题解 | #日期差值#

#include <stdbool.h>
#include <stdio.h>
bool learyear(int year)//判断是否闰年
{
    return (year%4==0&&year%100!=0||(year%400==0));
}
int numyear(int year)//判断每年多少天
{
    if(learyear(year))
        return 366;
    else
        return 365;
}
int main() {
    int year1,month1,day1,year2,month2,day2;
    scanf("%4d%2d%2d",&year1,&month1,&day1);
    scanf("%4d%2d%2d",&year2,&month2,&day2);
    int Days[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31},
                     {0,31,29,31,30,31,30,31,31,30,31,30,31}};
    int sum1=0,sum2=0;
  	//计算year1的天数
    for(int i=0;i<year1;i++)
    {
        sum1+=numyear(i);
    }
    for(int j=0;j<month1;j++)
    {
        sum1+=Days[learyear(year1)][j];
    }
    sum1+=day1;
	//计算year2的天数
    for(int i=0;i<year2;i++)
    {
        sum2+=numyear(i);
    }
    for(int j=0;j<month2;j++)
    {
        sum2+=Days[learyear(year1)][j];
    }
    sum2+=day2;
	//求出差值
    int num=sum2-sum1+1;
    if(num<0)
    {
        num=-num;
    }
    printf("%d",num);
    return 0;
}