Question

我无法正确计算蒙特卡洛Pi计划。基本上，pi目前只显示最多2个小数点，我觉得计算出错了，因为数字越高，pi计算越接近2.98-3.04。

我的代码粘贴在下面。

static void Main(string[] args)
{
    double n;
    double count;
    double c = 0.0;
    double x = 0.0, y = 0.0;
    double pi;
    string input;

    Console.WriteLine("Please input a number of dots for Monte Carlo to calculate pi.");
    input = Console.ReadLine();
    n = double.Parse(input);

    Random rand = new Random();


    for (int i = 1; i < n; i++ )
    {
        x = rand.Next(-1, 1);
        y = rand.Next(-1, 1);

        if (((x * x) + (y * y) <= 1))
            c++;
        pi = 4.0 * ( c / i );
        Console.WriteLine("pi: {0,-10:0.00} Dots in square: {1,-15:0} Dots in circle: {2,-20:0}", pi, i, c);
    }
}

Answer 1

这些电话

x = rand.Next(-1, 1);
y = rand.Next(-1, 1);

给你一个整数。但是你需要doubles：

x = rand.NextDouble() * 2 - 1;
y = rand.NextDouble() * 2 - 1;

Answer 2

随机数应该在0和1之间生成，而不是-1和1。使用此代码的固定版本作为＆＃34;神秘代码＆＃34;对于学生。

using System;

namespace mysCode
{
    class Program
    {
        static double euclideanDistance(double x1, double y1, double x2, double y2)
        {
            double dX = x2 - x1;
            double dY = y2 - y1;
            return Math.Sqrt(dX * dX + dY * dY);
        }

        static void Main(string[] args)
        {
            double n;            
            double c = 0.0;
            double x = 0.0, y = 0.0;
            double result;
            string input;
            Console.WriteLine("Quick, pick an integer");
            input = Console.ReadLine();
            n = double.Parse(input);
            Random rand = new Random();
            for (int i = 1; i <= n; i++)
            {
                x = rand.NextDouble();
                y = rand.NextDouble();
                if (euclideanDistance(x, y, 0, 0) <= 1)
                    c++;
                result = 4.0 * (c / i);                
                Console.WriteLine("Result: " + result);
            }
            Console.ReadKey();
        }

    }
}

覆盖率非常慢，在1M次迭代后得到3.14152314152314。

蒙特卡罗皮不准确

2 个答案: