编写一个程序，输出当年的月历。

显示效果如：

#include<stdio.h>
#include <stdlib.h>
#include <string.h>

typedef enum bool
{
	false, true
}bool;

bool isLeapYear(int year)//判断是否闰年
{
	if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0)
	{
		return true;
	}
	else
	{
		return false;
	}
}

int getMonthDays(int year, int month)//获得该月天数
{
	int days = 0;
	switch (month)
	{
	case 1:
	case 3:
	case 5:
	case 7:
	case 8:
	case 10:
	case 12:
		days = 31;
		break;
	case 4:
	case 6:
	case 9:
	case 11:
		days = 30;
		break;
	case 2:
		if (isLeapYear(year))
		{
			days = 29;
		}
		else
		{
			days = 28;
		}
		break;
	}
	return days;
}

void showMonth(int days, int startday)//输出月历
{
	const char weeks[][4] = { "SUN", "MON", "TUE", "WED", "THU", "FRI", "SAT" };
	for (int i = 0; i < 6; i++)
	{
		printf("%s ", weeks[i]);
	}
	printf("%s
", weeks[6]);

	for (int i = 1; i <= days; i++)
	{
		if (i == 1)
		{
			for (int j = 0; j < startday; j++)
			{
				printf("    ");
			}
			printf("%3d ", 1);
		}
		else
		{
			printf("%3d ", i);
		}
		if ((i + startday) % 7 == 0)
		{
			putchar('
');
		}
	}
	putchar('
');
}

int main(void)
{
	//以1970年1月1日(星期四)为基准
	int year, month, monthDays, startday, days;
	const int startYear = 1970;
	printf("请输入年月例：2015 8
");
	while (scanf("%d %d", &year, &month) != EOF)
	{
		monthDays = getMonthDays(year, month);
		days = 0;
		for (int i = startYear; i < year; i++)
		{
			if (isLeapYear(i))
			{
				days += 366;
			}
			else
			{
				days += 365;
			}
		}
		for (int i = 1; i < month; i++)
		{
			days += getMonthDays(year, i);
		}
		startday = (days + 4) % 7;
		//printf("%d
", startday);

		showMonth(monthDays, startday);
		printf("请输入年月例：2015 8
");
	}
	return 0;
}

输出：

请输入年月例：2015 8
2005 7
SUN MON TUE WED THU FRI SAT
                      1   2
  3   4   5   6   7   8   9
 10  11  12  13  14  15  16
 17  18  19  20  21  22  23
 24  25  26  27  28  29  30
 31
请输入年月例：2015 8