C#算法之实现阿姆斯特朗数

static void Main(string[] args)
        {
            int i = 0;
            int digitCount = 0;
            
            int[] digitArray = new int[10];
            double sum = 0;

            Console.Write("请输入验证的数字 : ");
            //1.输入数字
            int number = int.Parse(Console.ReadLine());
            //2.用临时变量保存该数字
            int temporaryNumber = number;
            //3.保存数字的总数以及每个数字存储在数字数组中
            while (number > 0)
            {
                digitArray[i++] = number % 10;
                number = number / 10;
                digitCount++;
            }
            //4.计算结果
            for (i = 0; i < digitCount; i++)
                //计算数字的次数幂
                sum += Math.Pow(digitArray[i], digitCount);

            //5.判断结果
            if (sum == temporaryNumber)
                Console.WriteLine($"{temporaryNumber} 属于阿姆斯特朗数");
            else
                Console.WriteLine($"{temporaryNumber} 不属于阿姆斯特朗数");
            Console.ReadLine();
        }

		static void Main(string[] args)
        {
            Console.Write("输入起始数字 : ");
            int StartNumber = int.Parse(Console.ReadLine());
            Console.Write("输入结束数字 : ");
            int EndNumber = int.Parse(Console.ReadLine());
            Console.WriteLine($"阿姆斯特朗数区间为: {StartNumber} ~ {EndNumber} 的结果为 : ");
            for (int i = StartNumber; i <= EndNumber; i++)
            {
                if (IsArmstrongNumber(i))
                    Console.WriteLine(i);
            }
            Console.ReadLine();
        }

        static bool IsArmstrongNumber(int number)
        {
            int sum = 0;
            int temporaryNumber = number;
            int temp = 0;
            int length = number.ToString().Length;
            while (number != 0)
            {
                temp = number % 10;
                number = number / 10;
                sum += (int)Math.Pow(temp, length);
            }

            if (sum == temporaryNumber)
                return true;
            else
                return false;
        }