题解 | 日期差值

#include <iostream>
using namespace std;
int mon[2000]={0,31,28,31,30,31,30,31,31,30,31,30,31,0};
bool run(int year){
    return year % 4 == 0 && year%100 != 0 || year % 400 == 0;
}

int main() {
    int a, b;
    while (cin >> a >> b) { // 注意 while 处理多个 case
        int res = 1,d=a%100,m=a/100%100,y=a/10000;
        int d1=b%100,m1=b/100%100,y1=b/10000;
        // cout<<y<<" "<<m<<" "<<d;
        while(d!=d1 || m!=m1 || y!=y1){
            if(!run(y))mon[y] = 28;
			else mon[2] = 29;
            res++;
            d++;
            if(d>mon[m]){
                m++;
                if(m==13){
                    y++;
                    m = 1;
                    if(!run(y))mon[y] = 28;
					else mon[2] = 29;
                }
                d=1;
            }
        //    cout<<y<<" "<<m<<" "<<d<<"--"<<res<<endl;
        } cout<<res<<endl;

        
        
    }
}
// 64 位输出请用 printf("%lld")