c#输出2023年的日历，每一周为一行。

Question

c#，设计控制台应用程序，输出2023年的日历，每一周为一行。
输出如下：

Answer 1

根据你给出的图显示，加入输入验证，实现如下，望采纳谢谢：

    internal class Program
    {

        public static void Main(string[] args)
        {
            Console.Write("请输入年份：");

            //输入验证
            int iYear;
            String strResult = Console.ReadLine();
            while (!Int32.TryParse(strResult, out iYear))
            {
                Console.WriteLine("请输入正确的年份！");
                Console.Write("\n");
                Console.Write("请输入年份：");
                strResult = Console.ReadLine();
            }

            bool bEstimate = LeapYear(iYear); // 判断是否闰年
            PrintDateOfYear(bEstimate, iYear); // 打印日期
            Console.ReadLine();
        }

        // 打印日历表
        private static void PrintDateOfYear(bool bEstimate,int iYear)
        {
            for (int month = 1; month <= 12; month++)
            {
                int j;
                int day = DayOfMonth(bEstimate, month);//判断月份的天数
                int blank = GetWeekByDay(iYear, month, 1);//计算空格

                Console.WriteLine(string.Format("{0}年{1}月", iYear, month));
                Console.Write(string.Format("{0}\t{1}\t{2}\t{3}\t{4}\t{5}\t{6}\n", "周日","周一","周二","周三","周四","周五","周六"));

                for (j = 1; j <= blank; j++)//添加空格
                {
                    Console.Write("\t");
                }

                for (int i = 1; i <= day; i++)
                {
                    int a = 8 - j;
                    if ((i + 7 - a) % 7 == 0 || i == day)
                    {
                        Console.Write(i + "\n");
                    }
                    else
                    {
                        Console.Write(i + "\t");
                    }
                }
                Console.Write("\n");

            }
        }

        private static int DayOfMonth(bool Estimate, int month)//判断指定月份的天数
        {
            int day;
            switch (month)
            {
                case 1: case 3: case 5: case 7: case 8: case 10: case 12:
                    day = 31;
                    break;
                case 4: case 6: case 9: case 11:
                    day = 30;
                    break;
                case 2:
                    if (Estimate)
                        day = 29;
                    else
                        day = 28;
                    break;
                default:
                    day = 0;
                    break;
            }

            return day;
        }

        // 判断是否为闰年
        private static bool LeapYear(int year)
        {
            if ((year % 400 == 0) || ((year % 4 == 0) && (year % 100 != 0))) return true;
            
            return false;
        }

        // 根据年月日计算星期数
        private static int GetWeekByDay(int year, int month, int day)
        {
            return (int)new DateTime(year, month, day).DayOfWeek;
        }
    }