题解 | 日期累加

#include <iostream>
using namespace std;
int days[] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
int main() {
    int year, month, day, n, cnt;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> year >> month >> day >> cnt;
        while (cnt != 0) {
            if (year % 400 == 0 || (year % 4 == 0 && year % 100 != 0))days[2] = 29;
            else days[2] = 28;
            int use = days[month] - day;
            if (use >= cnt) {
                day += use;
                day = cnt;
                break;
            }
            else {
                cnt -= use;
                if (month <= 11) {
                    ++month;
                    day = 0;
                }
                else {
                    ++year;
                    month = 1;
                    day = 0;
                }
            }
        }
        printf("%04d-%02d-%02d\n", year, month, day);
    }
}
// 64 位输出请用 printf("%lld")

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