题解 | #日期累加#日期累加思路简单解法

#include <iostream>
using namespace std;

bool IsLeapYear(int year) {
    return ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0);
}

int main() {
    int m;
    cin >> m;
    int year, month, day, daysum;
    while ((m--) &&
            (cin >> year >> month >> day >> daysum)) { // 注意 while 处理多个 case
	  //先给出每个月的日期数组
            int monthday[13] = {-1, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
	  //将当前的日期的day和要加的日期加在一起
        daysum += day;
        while (daysum > monthday[month]) {
            //若是闰年就是29否则28
            if (IsLeapYear(year)) {
                monthday[2] = 29;
            }
            else {
                monthday[2] = 28;
            }
		  //减点当前月的日期，然后month++
            daysum -= monthday[month];
            month++;
		  //如果month加到了12以上则将year++,将month重置为1
            if (month == 13) {
                year++;
                month = 1;
            }
        }


        printf("%04d-%02d-%02d\n", year, month, daysum);

    }
}
// 64 位输出请用 printf("%lld")