题解 | 打印日期

#include <iostream>

using namespace std;

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

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

int main() {
	int y, d;
	while (cin >> y >> d) {
		
		if (is_run(y)) days[2] = 29;
		else days[2] = 28;

		int m = 1;
		while (d > days[m]) {
			d -= days[m];
			m++;
		}

		printf("%d-%02d-%02d\n", y, m, d);
	}
	return 0;
}