题解 | #日期差值#

//繁琐但直接的思路
#include<iostream>
using namespace std;

int yy[2][12]={                                 //每月天数
        {31,28,31,30,31,30,31,31,30,31,30,31},
        {31,29,31,30,31,30,31,31,30,31,30,31}
    };

bool isrun(int year){                             //判断闰年
    return ((year%4==0&&year%100!=0)||year%400==0);
}
int rundays(int year){                          //对应天数
    if(isrun(year))return 366;
    else return 365;
}
bool compare(int year1,int month1,int day1,int year2,int month2,int day2){  //比较日期前后
    if(year1>year2||(year1==year2&&month1>month2)||(year1==year2&&month1==month2&&day1>day2))
        return false;
    else return true;
}
int main(){
    char s1[9],s2[9];
    int year1,month1,day1,year2,month2,day2;
    while(cin>>s1>>s2){
        int number=1;                //数据处理（直接用int做除法应该更简单）
        year1=(s1[0]-'0')*1000+(s1[1]-'0')*100+(s1[2]-'0')*10+(s1[3]-'0');
        month1=(s1[4]-'0')*10+(s1[5]-'0');
        day1=(s1[6]-'0')*10+(s1[7]-'0');
        
        year2=(s2[0]-'0')*1000+(s2[1]-'0')*100+(s2[2]-'0')*10+(s2[3]-'0');
        month2=(s2[4]-'0')*10+(s2[5]-'0');
        day2=(s2[6]-'0')*10+(s2[7]-'0');
        
        if(!compare(year1,month1,day1,year2,month2,day2)) {  //较小日期放到前面
            int t;
            t=year1;
            year1=year2;
            year2=t;
            t=month1;
            month1=month2;
            month2=t;
            t=day1;
            day1=day2;
            day2=t;
        }
                                          //分情况从第一个日期开始累计
        if(year1!=year2){                  //不同年
            int run=isrun(year1);          
            number+=yy[run][month1-1]-day1;
            while(month1<12)number+=yy[run][month1++];
            year1++;
            while(year1<year2){
                number+=rundays(year1++);
            }
            run=isrun(year2);
            while(month2>1)number+=yy[run][(month2--)-2];
            number+=day2;
        }
        else if(month1!=month2){  //同年不同月
            int run=isrun(year1);
            number+=yy[run][month1-1]-day1;
            while(month1+1<month2)number+=yy[run][month1++];
            number+=day2;
        }      
        else number+=(day2-day1);   //同年同月不同日
        
        cout<<number<<endl;
        }
    
    return 0;
}