题解 | #日期累加#

#include <iostream>
using namespace std;

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

bool isLeap(int year) {
    if (year % 400 == 0 || year % 4 == 0 && year % 100 != 0) {
        return true;
    }
    return false;
}

int main() {
    int m;
    while (scanf("%d", &m) != EOF) { // 注意 while 处理多个 case
        for (int j = 0; j < m; j++) { // m个日期
            int year, month, day, num;
            scanf("%d%d%d%d", &year, &month, &day, &num);
            if (isLeap(year)) {
                days[2] = 29;
            } else {
                days[2] = 28;
            }
            for (int i = 0; i < num; i++) {
                day++;
                if (day > days[month]) {
                    month++;
                    day = 1;
                }
                if (month > 12) {
                    year++;
                    if (isLeap(year)) { //年份改变了要看看是不是闰年
                        days[2] = 29;
                    } else {
                        days[2] = 28;
                    }
                    month = 1;
                }
            }
            printf("%d-%02d-%02d\n", year, month, day);
        }

    }
}
// 64 位输出请用 printf("%lld")