题解 | #日期累加#

#include <iostream>
using namespace std;

int main() {
    int N = 0;
    cin >> N;
    while (N--) {
        int monthall[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        int year, month, day, n;
        cin >> year >> month >> day >> n;
        day += n;
        if (day > 59 && (year % 4 == 0 && year % 100 != 0) || ( year  % 400 == 0)) {
            monthall[2] = 29;
        }
        while (day > monthall[month]) {

            day -= monthall[month];
            month++;
            if (month > 12) {
                month = 1;
                year++;
                if (day > 59 && (year % 4 == 0 && year % 100 != 0) || ( year  % 400 == 0)) {
                    monthall[2] = 29;
                }
                else {
                monthall[2] = 28;
                }
            }
        }
        if (day >= 10 && month >= 10)

            printf("%d-%d-%d", year, month, day);

        else if (day < 10 && month >= 10)

            printf("%d-%d-0%d", year, month, day);

        else if (day >= 10 && month < 10)

            printf("%d-0%d-%d", year, month, day);

        else

            printf("%d-0%d-0%d", year, month, day);

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