题解 | #日期差值#

#include <iostream>
#include <functional>
#include <iostream>
using namespace std;
int isPyear(int y){
    return (y%4==0 && y%100!=0)||(y%400==0);
}

int main() {
int daysEach[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 d1,d2;
cin>>d1>>d2;
if(d1>d2){
    int tem=d1;
    d1=d2;
    d2=tem;
}
int minus_day=(d2-d1)%100;
if(minus_day>=70) minus_day=minus_day-100;//day of d2 is lower than that of d1
int minus_mon=(d2-d1)%10000/100;
if(minus_mon>=89) minus_mon-=100;
int d1_year=d1/10000;
int d1_mon=d1/100%100;
if(minus_mon<0){
    int j=-minus_mon;
    while(j>0){
    	d1_mon--;
        minus_day-=daysEach[isPyear(d1_year)][d1_mon];
        j--;
    }
}
else {
    while(minus_mon>0){
        minus_day+=daysEach[isPyear(d1_year)][d1_mon];
        minus_mon--;
        d1_mon--;
    }
}
int d2_year=d2/10000;
int count=1;
while(d1_year<d2_year){
    if(isPyear(d1_year)) count+=366;
    else count+=365;
    d1_year++;
}
count+=minus_day;
cout<<count;
}