题解 | 日期差值

#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<vector<int>> a = {
    {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}
};

bool isLeapyear(int yr){
    return ((yr%4==0&&yr%100!=0)||yr%400==0)?true:false;
}
//计算到较小年份1月1日的距离。
int daystomin(int minyear,int year,int month,int day){
    int count=0;
    while(year>minyear){
        count+=(isLeapyear(minyear)?366:365);
        minyear++;
    }
    for(int i=1;i<month;i++){
        count+=(isLeapyear(year)?a[1][i]:a[0][i]);
    }
    count+=day;
    count--;
    return count;
}

int main() {
    string a,b;
    while(cin>>a>>b){
        int year1,year2,month1,month2,day1,day2;
        year1=stoi(a.substr(0,4));
        year2=stoi(b.substr(0,4));
        int minyear = min(year1,year2);
        month1 = stoi(a.substr(4,2));
        month2 = stoi(b.substr(4,2));
        day1 = stoi(a.substr(6,2));
        day2 = stoi(b.substr(6,2));
        int distance1=daystomin(minyear,year1,month1,day1);
        int distance2=daystomin(minyear,year2,month2,day2);
        int result = abs(distance1-distance2)+1;
        cout<<result<<endl;
    }
    
}