[C++知识库]C语言实现:输出明天的日期

#include<stdio.h>
struct data {
	int year, month, day;
};

int main() {
	struct data today;
	scanf("%d.%d.%d", &today.year, &today.month, &today.day);
	int flag = (today.year % 4 == 0 && today.year % 100 != 0) || today.year % 400 == 0;//flat==1表示,今年为闰年
	today.day++;
	switch (today.month) {
		case 1:case 3:case 5:case 7:case 8:case 10:
		case 12: {
			if (today.day == 32) {
				today.month++;
				today.day = 1;
			}
			break;
		}
		case 4:case 6:case 9:case 11: {
			if (today.day == 31) {
				today.month++;
				today.day = 1;
			}
			break;
		}
		case 2: {
			if (flag == 1 && today.day == 30) {
				today.month++;
				today.day = 1;
			}
			if (flag == 0 && today.day == 29) {
				today.month++;
				today.day = 1;
			}
		}
	}
	if(today.month == 13){
		today.year++;
		today.month=1;
	}
	printf("%04d.%02d.%02d",today.year,today.month,today.day);
	return 0;
}