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基本思路：算year1到year2的总共天数，然后减去多算的天数，多算的天数可以通过参考“已知日期求该日是这年的第几天”求出。

 

#include<bits/stdc++.h>

using namespace std;

int main(){
    int year1,month1,day1;
    int year2,month2,day2;
    int f[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
    scanf("%4d%2d%2d",&year1,&month1,&day1);
    scanf("%4d%2d%2d",&year2,&month2,&day2);

    int sum=0,sum1=day1,sum2=day2;

    for(int i=year1;i<=year2;i++){
        if(i%400==0||(i%4==0&&i%100!=0)) sum+=366;
        else sum+=365;
    }
    if(year1%400==0||year1%4==0&&year1%100!=0) f[2]=29;
    else f[2]=28;

    for(int i=1;i<month1;i++){
        sum1+=f[i];
    }
    //cout<<sum1;
    if(year2%400==0||year2%4==0&&year2...