题解 | #日期差值#格式化读入

#include <iostream>
using namespace std;

int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

char a[9], b[9];

int main() {
    while (scanf("%s%s", a, b) != EOF) { // 注意 while 处理多个 case   
		int year1, year2;
        int month1, month2;
        int day1, day2;
        sscanf(a,"%4d%2d%2d",&year1,&month1,&day1);//格式化读入
        sscanf(b,"%4d%2d%2d",&year2,&month2,&day2);
        
        //把year1变成小的日期
        if (year1 < year2 || year1 == year2 && month1 < month2
                || year1 == year2 && month1 == month2 && day1 < day2) { //不用交换的情况

        } else {
            int t = year1;
            year1 = year2;
            year2 = t;
            t = month1;
            month1 = month2;
            month2 = t;
            t = day1;
            day1 = day2;
            day2 = t;
        }

        int count = 0;
        while (year1 != year2 || month1 != month2 || day1 != day2) {
            day1++;
            count++;
            if (day1 > days[month1]) {
                month1++;
                day1 = 1;
            }
            if (month1 > 12) {
                year1++;
                month1 = 1;
            }
        }
        printf("%d\n", count + 1);
    }
}
// 64 位输出请用 printf("%lld")

采用格式化的读入节省了很多代码