题解 | #计算日期到天数转换#

#include <stdio.h>
//第一种简单数组方法
int main()
{
	int year;
	int month;
	int day;
	scanf("%d %d %d", &year, &month, &day);
	int sum = day;

//闰年数组：
	int leap_year[12] = { 31,29,31,30,31,30,31,31,30,31,30,31 };
	                 //   1  2  3  4  5  6  7  8  9  10  11 12

//平年数组：
	int common_year[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
	                //       1  2  3  4  5  6  7  8  9 10 11 12
//闰年情况：
	if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
	{
		if (month == 1)
		{
			sum = day;
		}
		else
		{
			int j = 0;

			for (int i = 1; i < month; i++)
			{
				sum = sum + leap_year[j++];
			}
		}

	}
//平年情况：
	else
	{
		if (month == 1)
		{
			sum = day;
		}
		else
		{
			int j = 0;

			for (int i = 1; i < month; i++)
			{
				sum = sum + common_year[j++];
			}
		}

	}
	printf("%d", sum);
	return 0;
}
//第二种暴力方法：
int main()
{
	int year;
	int month;
	int day;
	scanf("%d %d %d", &year, &month, &day);
	int sum = 0;
	//1 3 5 7 8 10 12 ——31天
	//4 6 9 11        ——30天
	//闰年2月         ——29天
	//平年2月         ——28天
	//如果year是闰年 2000 11 15
	if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
	{
		switch (month)
		{
		case 1:
			sum = day;
			break;
		case 2:
			sum = 31 + day;
			break;
		case 3:
			sum = 31 + 29 + day;
			break;
		case 4:
			sum = 31 + 29 + 31 + day;
			break;
		case 5:
			sum = 31 + 29 + 31 + 30 + day;
			break;
		case 6:
			sum = 31 + 29 + 31 + 30 + 31 + day;
			break;
		case 7:
			sum = 31 + 29 + 31 + 30 + 31 + 30 + day;
			break;
		case 8:
			sum = 31 + 29 + 31 + 30 + 31 + 30 + 31 + day;
			break;
		case 9:
			sum = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + day;
			break;
		case 10:
			sum = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + day;
			break;
		case 11:
			sum = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + day;
			break;
		case 12:
			sum = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + day;
			break;
		}
	}
	else
	{
		switch (month)
		{
		case 1:
			sum = day;
			break;
		case 2:
			sum = 31 + day;
			break;
		case 3:
			sum = 31 + 28 + day;
			break;
		case 4:
			sum = 31 + 28 + 31 + day;
			break;
		case 5:
			sum = 31 + 28 + 31 + 30 + day;
			break;
		case 6:
			sum = 31 + 28 + 31 + 30 + 31 + day;
			break;
		case 7:
			sum = 31 + 28 + 31 + 30 + 31 + 30 + day;
			break;
		case 8:
			sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + day;
			break;
		case 9:
			sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + day;
			break;
		case 10:
			sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + day;
			break;
		case 11:
			sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + day;
			break;
		case 12:
			sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + day;
			break;
		}
	}
	printf("%d", sum);
	return 0;
}