题解 | #日期差值#

#include <bits/stdc++.h>
using namespace std;
int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int leap(int year){
	if((year%4==0 && year%100!=0) || year%400==0) return 1;
	return 0;
}

int main(){
	int day1,day2,mon1,mon2,year1,year2;
	scanf("%4d%2d%2d",&year1,&mon1,&day1);
	scanf("%4d%2d%2d",&year2,&mon2,&day2);
	//算出距离0000-00-00的绝对时间，再相减即可
	int sum1=0,sum2=0;
	for(int yy=0;yy<year1;yy++){
		if(leap(yy)) sum1+=366;
		else sum1+=365;
	} 
	if(leap(year1)) day[2]=29;
		else day[2]=28;
	for(int mm=1;mm<mon1;mm++){
		sum1+=day[mm];
	}
	sum1+=day1;
	
	for(int yy=0;yy<year2;yy++){
		if(leap(yy)) sum2+=366;
		else sum2+=365;
	} 
	if(leap(year2)) day[2]=29;
		else day[2]=28;
	for(int mm=1;mm<mon2;mm++){
		sum2+=day[mm];
	}
	sum2+=day2;
	cout<<abs(sum1-sum2)+1<<endl;
	return 0;
}