循环嵌套练习题

            //BOSS://让用户输入一个奇数，打印菱形，最长的行内容个数为用户输入的个数，并且由英文字母拼接而成//比如用户输入了7//    A//   ABA//  ABCBA// ABCDCBA//  ABCBA//   ABA//    A//1、接收并判断用户输入的是不是数字  try{#region  解法一//Console.Write("请输入一个奇数：");//int a = Convert.ToInt32(Console.ReadLine());//if (a % 2 != 0)//{//    for (int i = 1; i <= (a + 1) / 2; i++)//    {//        for (int b = 1; b <= ((a + 1) / 2 - i); b++)//        {//            Console.Write(" ");//        }//        char c = 'A';//        for (int d = 1; d < i; d++)//        {//            Console.Write(c);//            c++;//        }//        for (int e = 1; e <= 26; e++)//        {//            Console.Write(c);//            if (c == 'A')//            {//                break;//            }//            c--;//        }//        Console.WriteLine();//    }//    for (int i = 1; i < (a + 1) / 2; i++)//    {//        for (int b = 1; b <= i; b++)//        {//            Console.Write(" ");//        }//        char c = 'A';//        for (int d = 1; d < (a + 1) / 2 - i; d++)//        {//            Console.Write(c);//            c++;//        }//        for (int e = 1; e <= 26; e++)//        {//            Console.Write(c);//            if (c == 'A')//            {//                break;//            }//            c--;//        }//        Console.WriteLine();//    }//}#endregion#region 解法二 上半部分               Console.Write("请输入一个奇数：");int a = Convert.ToInt32(Console.ReadLine());if (a % 2 != 0)//是奇数，执行这个if里面的代码
                {for (int i = 1; i <= (a + 1) / 2; i++)//上半部分行数，（a+1）/2 代表 上半部分需要打印的行数。
                    {char ch = 'A';string end = "";int b = ((i * 2 - 1) + 1) / 2 - 1;//开始 -- 的数值bool isok = false;int count = 0;for (int j = 1; j <= ((a + 1) / 2) - i; j++)//拼接每行打印的空格数
                        {end += " ";}for (int j = 1; j <= i * 2 - 1; j++)//拼接每行打印的字母数
                        {end += ch;if (count == b)//判断是不是该 -- 了
                            {isok = true;}if (isok)//满足条件，执行这个 if 里面的代码
                            {if (ch == 'A'){ch = 'Z';}else//条件不成立
                                {ch--;}}else{if (ch == 'Z'){ch = 'A';}else{ch++;}count++;}}Console.WriteLine(end);}}#endregion#region 解法二 下半部分for (int i = 1; i < (a + 1) / 2; i++){char ch = 'A';string end1 = "";int b = ((a - i * 2) + 1) / 2 - 1;int count = 0;bool isok = false;for (int j = 1; j <= i; j++){end1 += " ";}for (int j = 1; j < (a - i * 2) + 1; j++){end1 += ch;if (count == b){isok = true;}if (isok){if (ch == 'A'){ch = 'Z';}else{ch--;}}else{if (ch == 'Z'){ch = 'A';}else{ch++;}count++;}}Console.WriteLine(end1);}#endregion}catch{Console.WriteLine("输入有误！");}Console.ReadLine();