题解 | #日期累加#

#include <cstdio>
#include <iostream>
using namespace std;

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

int main() {
    int m;
    int _day[2][13] = {{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}
    };
    scanf("%d", &m);
    for (int i = 0; i < m ; ++i) {
        int year, month, day, needSumDay;
        scanf("%d %d %d %d", &year, &month, &day, &needSumDay);
        //1.当需要累加的天数大于当月剩余的天数，则月份++，并减去当月剩余天数得到新的需要累加的天数
	  //2.当月份为12月时，则需要将年份++
        while (needSumDay > _day[IsLeapYear(year)][month] - day) {
            needSumDay -= _day[IsLeapYear(year)][month] - day;
            day=0;
            if (month < 12) {
                ++month;
            }else {
                month = 1;
                ++year;
            }
        }
        printf("%04d-%02d-%02d\n",year,month,needSumDay);

    }
    return 0;
}