函数附加练习1

1、写两个函数，分别求两个整数的最大公约数和最小公倍数，用主函数调用这两个函数，并输出结果。两个整数由键盘输入。

主要代码：

        static void Main(string[] args)
        {
            Program function = new Program();
            Console.Write("请输入：");
            double num1 = double.Parse(Console.ReadLine());
            Console.Write("请输入：");
            double num2 = double.Parse(Console.ReadLine());
            double num = function.divisor(num1, num2);
            double Num = function.multiple(num1, num2);
            Console.WriteLine("最大公约数为：{0}，最小公倍数为：{1}。", num, Num);
            Console.ReadLine();
        }
        /// <summary>
        /// 最大公约数
        /// </summary>
        /// <param name="num1"></param>
        /// <param name="num2"></param>
        /// <returns></returns>
        public double divisor(double num1, double num2)
        {
            double temp = 0;
            double num = 0;
            if (num1 < num2)
            {
                temp = num1;
                num1 = num2;
                num2 = temp;
            }
            while (num2 != 0)
            {
                num = num1 % num2;
                num1 = num2;
                num2 = num;
            }
            return num1;
        }
        /// <summary>
        /// 最小公倍数
        /// </summary>
        /// <param name="num1"></param>
        /// <param name="num2"></param>
        /// <returns></returns>
        public double multiple(double num1, double num2)
        {
            double num = 0;
            Program method = new Program();
            double Num = method.divisor(num1, num2);
            num = num1 * num2 / Num;
            return num;
        }

结果：

2、求方程ax2+bx+c=0的根，用3个函数分别求当b2-4ac大于0、等于0和小于0时的根并输出，从主函数输入a，b，c的值。

主要代码：

        static void Main(string[] args)
        {
            Program function = new Program();
            Console.Write("请输入a：");
            double a = double.Parse(Console.ReadLine());
            Console.Write("请输入b：");
            double b = double.Parse(Console.ReadLine());
            Console.Write("请输入c：");
            double c = double.Parse(Console.ReadLine());
            if (b * b - 4 * a * c > 0)
            {
                double x1 = function.equation0(a, b, c);
                double x2 = function.equation1(a, b, c);
                Console.WriteLine("x1={0:f2},x2={1:f2}", x1, x2);
            }
            else if (b * b - 4 * a * c == 0)
            {
                double x = function.equation2(a, b, c);
                Console.WriteLine("x={0:f2}", x);
            }
            else
                function.equation();
            Console.ReadLine();         
        }      
        /// <summary>
        /// 求一元二次方程，△>0
        /// </summary>
        /// <param name="a"></param>
        /// <param name="b"></param>
        /// <param name="c"></param>
        /// <returns></returns>
        public double equation0(double a, double b, double c)
        {
            double D = b * b - 4 * a * c;
            double x = (((-b) + Math.Sqrt(D)) / (2 * a));
            return x;
        }
        public double equation1(double a, double b, double c)
        {
            double D = b * b - 4 * a * c;
            double x = (((-b) - Math.Sqrt(D)) / (2 * a));
            return x;
        }
        /// <summary>
        /// 求一元二次方程，△=0
        /// </summary>
        /// <param name="a"></param>
        /// <param name="b"></param>
        /// <param name="c"></param>
        /// <returns></returns>
        public double equation2(double a, double b, double c)
        {
            double D = b * b - 4 * a * c;
            double x = (((-b) + Math.Sqrt(D)) / (2 * a));
            return x;
        }
        /// <summary>
        /// 求一元二次方程，△<0
        /// </summary>
        public void equation()
        {
            Console.WriteLine("此方程无实根！");
        }

        static void Main(string[] args)
        {
            //△>0的另一种写法。
            Program function = new Program();
            double x1;
            double x2;
            function.equation(out x1, out x2);
            Console.WriteLine("x1={0:f2},x2={1:f2}", x1, x2);
            Console.ReadLine();
        }
        public void equation(out double x1, out double x2)
        {
            Console.Write("请输入a：");
            double a = double.Parse(Console.ReadLine());
            Console.Write("请输入b：");
            double b = double.Parse(Console.ReadLine());
            Console.Write("请输入c：");
            double c = double.Parse(Console.ReadLine());
            double D = b * b - 4 * a * c;
            x1 = (((-b) + Math.Sqrt(D)) / (2 * a));
            x2 = (((-b) - Math.Sqrt(D)) / (2 * a));
        }

结果：

1）△>0

2）△=0

3）△<0

3、写一个判断素数的函数，在主函数输入一个整数，输出是否为素数的信息。

主要代码：

        static void Main(string[] args)
        {
            Program method = new Program();
            Console.Write("请输入：");
            int a = int.Parse(Console.ReadLine());
            method.number(a);
            Console.ReadLine();
        }
        public void number(int n)
        {
            int count = 0;
            for (int i = 1; i <= n; i++)
            {
                if (n % i == 0)
                {
                    count++;
                }
            }
            if (count == 2)
                Console.WriteLine("你输入的是素数！");
            else
                Console.WriteLine("你输入的不是素数！");
        }

结果：

4、写一个函数，是给定的一个nxn的二维整形数组转置

主要代码：

        static void Main(string[] args)
        {
            //法一
            Program method = new Program();
            Console.Write("请输入维数n：");
            int n = int.Parse(Console.ReadLine());
            double[,] m = new double[n, n];
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    Console.Write("请输入：");
                    m[i, j] = double.Parse(Console.ReadLine());
                }
            }
            Console.WriteLine("原数组：");
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    Console.Write(m[i, j] + "	");
                }
                Console.WriteLine();
            }
            Console.WriteLine("转置数组：");
            method.array(m, n);
            Console.ReadLine();
        }
        public void array(double[,] m, double n)
        {
            double temp = 0;
            for (int i = 0; i < n; i++)
            {
                for (int j = i + 1; j < n; j++)
                {
                    temp = m[i, j];
                    m[i, j] = m[j, i];
                    m[j, i] = temp;
                }
            }
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    Console.Write(m[i, j] + "	");
                }
                Console.WriteLine();
            }
        }

        static void Main(string[] args)
        {
            //法二
            Program method = new Program();
            Console.Write("请输入维数n：");
            int n = int.Parse(Console.ReadLine());
            double[,] m = new double[n, n];
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    Console.Write("请输入：");
                    m[i, j] = double.Parse(Console.ReadLine());
                }
            }
            Console.WriteLine("原数组：");
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    Console.Write(m[i, j] + "	");
                }
                Console.WriteLine();
            }
            Console.WriteLine("转置数组：");
            double[,] m1 = method.array(m, n);
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    Console.Write(m1[i, j] + "	");
                }
                Console.WriteLine();
            }
            Console.ReadLine();
        }
        public double[,] array(double[,] m, int n)
        {
            double temp = 0;
            for (int i = 0; i < n; i++)
            {
                for (int j = i + 1; j < n; j++)
                {
                    temp = m[i, j];
                    m[i, j] = m[j, i];
                    m[j, i] = temp;
                }
            }
            return m;
        }

结果：