1、求阶乘

            Console.WriteLine("输入一个数");
            int n = Convert.ToInt32(Console.ReadLine());
            int s = 1;

            for (int i = 1; i <= n; i++)
            {
                s = s * i;
            }
            Console.WriteLine("结果:" + s);

例:求5!

 

 

2、求阶乘的和

            Console.WriteLine("输入一个数:");
            int n = Convert.ToInt32(Console.ReadLine());
            int a = 0;

            for (int i = 1; i <= n; i++)
            {
                int s = 1;
                for (int j = 1; j <= i; j++)
                {
                    s = s * j;
                }
                a = a + s;
            }

            Console.WriteLine("结果:" + a);

例:5!+4!+3!+2!+1!=

 

 

3、找出100以内质数,并求和

            int sum = 0;
            for (int i = 2; i <= 100; i++)
            {
                int a = 0;

                for (int j = 1; j <= i; j++)
                {
                    if (i % j == 0)
                    {
                        a++;
                    }
                }
                if (a == 2)
                {
                    sum = sum + i;

                    Console.Write(i + "\t");
                }
            }
            Console.Write("总和:" + sum);

 

 

 

4、100元购物券买香皂(2元)、牙刷(5元)、洗发水(15元),每样至少买一个,正好花光,求所有可能

            int n = 1;
            for (int i = 1; 2 * i < 100; i++)
            {
                for (int j = 1; 5 * j < 100; j++)
                {
                    for (int k = 1; 15 * k < 100; k++)
                    {
                        if (2 * i + 5 * j + 15 * k == 100)
                        {
                            Console.WriteLine(n);
                            Console.WriteLine("香皂:" + i);
                            Console.WriteLine("牙刷:" + j);
                            Console.WriteLine("洗发水:" + k);
                            n++;
                        }
                    }
                }
            }

 

5、公鸡2文,母鸡1文,小鸡半文,每种至少买一只,100文钱买100只鸡,求所有可能

            int n = 1;
            for (int i = 1; 2 * i < 100; i++)
            {
                for (int j = 1; j < 100; j++)
                {
                    for (int k = 1; 0.5 * k < 100; k++)
                    {
                        if (2 * i + j + 0.5 * k == 100 && i + j + k == 100)
                        {
                            Console.WriteLine(n);
                            Console.WriteLine("公鸡{0}只,母鸡{1}只,小鸡{2}只", i, j, k);  //"{0}",占位符,必须从0开始
 
                            n++;
                        }
                    }
                }
            }

 

 

posted on 2014-08-19 16:50  幻影K  阅读(263)  评论(0编辑  收藏  举报