方法一：

题目要求打印12月的日历，第一行一定是一样的，我们放在一个字符串s中，然后就只剩下两部分需要考虑了，一是从星期几开始，二是一个月要打多少天。

第一部分可以这样考虑，题目中已经给出了9月的日历，可以发现是从星期一开始的，也就是没有空格，这点性质很好，我们接下来这样考虑：如果比到9.1这一天每多一天，就应该多一个空格，当然，因为日历会换行，最多也就是多六个空格，那么就有对应六种情况，分别是多：0、1、2、3、4、5、6个空格，同样的，每当比到9.1这一天少一天，就会有6、5、4、3、2、1、0个空格（因为如果比九月少会从上一月的最后一天开始），那么很容易想到对7取余，如果小于0就加上7，刚好是要打印的空格数。我们用一个前缀和数组处理某月到9.1隔了多少天。

接下来就只剩下打印日期了，我们把每个月的日期存下来，打印m[i]次即可，每当空格加上当前打印天数是7的倍数，就换行，当然，要注意空格以及向右对齐。

#include<iostream>
#include<string>

using namespace std;

int main()
{
	string s = "MON TUE WED THU FRI SAT SUN";
	int m[] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
	int sum[13] = { 0 };
	for (int i = 1; i <= 12; i++)
	{
		sum[i] = sum[i - 1] + m[i];//前缀和数组，到某月的总共天数
	}
	int base = sum[8]+1;//到9.1号的总共天数
	int month;
	cin >> month;
	int space_num = (sum[month - 1] + 1 - base) % 7;//每月月初到9.1日期的间隔
	if (space_num < 0)//小于零说明月份比九月小
		space_num += 7;
	cout << s << endl;
	for (int i = 0; i < space_num; i++)//处理空格
	{
		cout << "    ";
	}
	int i = 0;
	while (i < m[month])
	{
		printf("%3d ", i + 1);//%3d表示宽度为3，不足用空格填充
		i++;
		if ((i+space_num) % 7 == 0)
			cout << endl;
	}
}

方法二：

打开你的日历，可以马上看到每一个月的天数alt 接下来打表即可。

#include<iostream>
#include<string>

using namespace std;

int main()
{
	string s = "MON TUE WED THU FRI SAT SUN";
	int m_d[] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
	string m[] = { "    ","  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 " };
	int space_num[] = { 0,2,5,5,1,3,6,1,4,0,2,5,0 };
	int month;
	cin >> month;
	cout << s << endl;
	for (int i = 0; i < space_num[month]; i++)
	{
		cout << m[0];
	}
	int i = 0;
	while (i < m_d[month])
	{
		cout << m[i+1];
		i++;
		if ((i + space_num[month]) % 7 == 0)
			cout << endl;
	}
}