Question

我对编程非常陌生并且在尝试使用Heron公式创建程序时遇到问题，计算了数字的平方根的5个近似值 i ＆lt; = 10.

公式为：

X _n + 1 =（ X _n +（ i / X _n ））/ 2

我提出了以下代码，但它没有工作，因为输出什么都没有。

SCRIPT_NAME=/status SCRIPT_FILENAME=/status REQUEST_METHOD=GET cgi-fcgi -bind -connect /var/run/php/php7.0-fpm.sock

我非常感谢与此问题相关的帮助或提示。

Answer 1

你的while语句后面有一个额外的分号，导致整个明显的身体被忽略：

    while (zahl <= 10);    // <-- REMOVE THIS SEMICOLON
    {
        j = 0;
        while (j++ < 5)
        {
            Console.WriteLine("{0,5}", root = (root + (zahl / root)) / 2);
        }
    }

Answer 2

我建议将while个圈转换为for个圈：

 static void Main(string[] args) {
   // for each number in [0..10] range
   for (double number = 1.0; number <= 10.0; ++number) {
     double root = number / 2.0; // <- initial (zeroth) root aproximation

     // we want to calculate 5 approximations:
     for (int app = 1; app <= 5; ++app) {
       root = (root + number / root) / 2.0;

       // string interpolation $"..{some value} .." can be very handy 
       Console.WriteLine($"sqrt({number}) == {root} (app #{i})");
     } 

     // Let's split output 
     Console.WriteLine(); 
   }
 }

小心

条件：while (zahl <= 10) {...}应在zahl内更改{...}以防止无限循环
分号：while (zahl <= 10);表示“zahl＆lt; = 10时不执行任何操作”会导致无限循环

结果：

sqrt(1) == 1.25 (approximation #1)
sqrt(1) == 1.025 (approximation #2)
sqrt(1) == 1.00030487804878 (approximation #3)
sqrt(1) == 1.00000004646115 (approximation #4)
sqrt(1) == 1 (approximation #5)

sqrt(2) == 1.5 (approximation #1)
sqrt(2) == 1.41666666666667 (approximation #2)
sqrt(2) == 1.41421568627451 (approximation #3)
sqrt(2) == 1.41421356237469 (approximation #4)
sqrt(2) == 1.41421356237309 (approximation #5)

sqrt(3) == 1.75 (approximation #1)
sqrt(3) == 1.73214285714286 (approximation #2)
sqrt(3) == 1.73205081001473 (approximation #3)
sqrt(3) == 1.73205080756888 (approximation #4)
sqrt(3) == 1.73205080756888 (approximation #5)

....

sqrt(9) == 3.25 (approximation #1)
sqrt(9) == 3.00961538461538 (approximation #2)
sqrt(9) == 3.00001536003932 (approximation #3)
sqrt(9) == 3.00000000003932 (approximation #4)
sqrt(9) == 3 (approximation #5)

sqrt(10) == 3.5 (approximation #1)
sqrt(10) == 3.17857142857143 (approximation #2)
sqrt(10) == 3.16231942215088 (approximation #3)
sqrt(10) == 3.16227766044414 (approximation #4)
sqrt(10) == 3.16227766016838 (approximation #5)

苍鹭在C＃中的配方

2 个答案: