如何使用库调用在 C# 中计算阶乘？

How can I calculate a factorial in C# using a library call?

提问by mmr

I need to calculate the factorial of numbers up to around 100! in order to determine if a series of coin flip-style data is random, as per this Wikipedia entry on Bayesian probability.As you can see there, the necessary formula involves 3 factorial calculations (but, interestingly, two of those factorial calculations are calculated along the way to the third).

我需要计算高达 100 左右的数字的阶乘！为了确定一系列硬币翻转式数据是否是随机的，根据维基百科关于贝叶斯概率的条目。正如您在那里看到的，必要的公式涉及 3 个阶乘计算（但有趣的是，其中两个阶乘计算是在第三个阶乘计算的过程中计算出来的）。

I saw this question here, but I'd think that integer is going to get blown out pretty quickly. I could also make a function that is more intelligent about the factorial calculation (ie, if I have 11!/(7!3!), as per the wiki example, I could go to (11*10*9*8)/3!), but that smacks of premature optimization to me, in the sense that I want it to work, but I don't care about speed (yet).

我在这里看到了这个问题，但我认为整数很快就会被炸毁。我还可以创建一个对阶乘计算更智能的函数（即，如果我有 11!/(7!3!)，根据 wiki 示例，我可以转到 (11*10*9*8)/ 3！），但这对我来说有点过早优化，因为我希望它工作，但我不关心速度（还）。

So what's a good C# library I can call to calculate the factorial in order to get that probability? I'm not interested in all the awesomeness that can go into factorial calculation, I just want the result in a way that I can manipulate it. There does not appear to be a factorial function in the Math namespace, hence the question.

那么什么是好的 C# 库，我可以调用它来计算阶乘以获得该概率？我对可以进入阶乘计算的所有令人敬畏的东西不感兴趣，我只想以一种我可以操纵它的方式得到结果。Math 命名空间中似乎没有阶乘函数，因此存在问题。

采纳答案by TrueWill

You could try Math.NET- I haven't used that library, but they do list Factorial and Logarithmic Factorial.

你可以试试Math.NET——我没有使用过那个库，但他们确实列出了 Factorial 和 Logarithmic Factorial。

回答by bobbymcr

There has been a previous questionon a similar topic. Someone there linked the Fast Factorial Functionsweb site, which includes some explanations of efficient algorithms and even C# source code.

已经有一个先前的问题上类似的话题。有人链接了Fast Factorial Functions网站，其中包括对高效算法的一些解释，甚至 C# 源代码。

回答by Kirk Broadhurst

Do you want to calculate factorials, or binomial coefficients?

你想计算阶乘还是二项式系数？

It sounds like you want to calculate binomial coefficients - especially as you mention 11!/(7!3!).

听起来您想计算二项式系数 - 特别是当您提到 11!/(7!3!) 时。

There may be a library that can do this for you, but as a (presumably) programmer visiting stack overflow there's no reason not to write one yourself. It's not too complicated.

可能有一个库可以为您执行此操作，但是作为访问堆栈溢出的（大概）程序员，没有理由不自己编写一个。这不是太复杂。

To avoid memory overflow, don't evaluate the result until all common factors are removed.

为避免内存溢出，在删除所有公因子之前不要评估结果。

This algorithm still needs to be improved, but you have the basis for a good algorithm here. The denominator values need to be split into their prime factors for the best result. As it stands, this will run for n = 50 quite quickly.

这个算法还需要改进，但是你这里有一个好的算法的基础。需要将分母值拆分为其主要因子以获得最佳结果。就目前而言，这将很快运行 n = 50。

float CalculateBinomial(int n, int k)
{
    var numerator = new List<int>();
    var denominator = new List<int>();
    var denominatorOld = new List<int>();

    // again ignore the k! common terms
    for (int i = k + 1; i <= n; i++)
        numerator.Add(i);

    for (int i = 1; i <= (n - k); i++)
    {
        denominator.AddRange(SplitIntoPrimeFactors(i));
    }

    // remove all common factors
    int remainder;                
    for (int i = 0; i < numerator.Count(); i++)
    {
        for (int j = 0; j < denominator.Count() 
            && numerator[i] >= denominator[j]; j++)
        {
            if (denominator[j] > 1)
            {
                int result = Math.DivRem(numerator[i], denominator[j], out remainder);
                if (remainder == 0)
                {
                    numerator[i] = result;
                    denominator[j] = 1;
                }
            }
        }
    }

    float denominatorResult = 1;
    float numeratorResult = 1;

    denominator.RemoveAll(x => x == 1);
    numerator.RemoveAll(x => x == 1);

    denominator.ForEach(d => denominatorResult = denominatorResult * d);
    numerator.ForEach(num => numeratorResult = numeratorResult * num);

    return numeratorResult / denominatorResult;
}

static List<int> Primes = new List<int>() { 2, 3, 5, 7, 11, 13, 17, 19, 
    23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 };

List<int> SplitIntoPrimeFactors(int x)
{
    var results = new List<int>();
    int remainder = 0;

    int i = 0;
    while (!Primes.Contains(x) && x != 1)
    {
        int result = Math.DivRem(x, Primes[i], out remainder);
        if (remainder == 0)
        {
            results.Add(Primes[i]);
            x = result;
            i = 0;
        }
        else
        {
            i++;
        }
    }
    results.Add(x);
    return results;
}

I can estimate n = 110, k = 50 (returns 6x10^31) but cannot run n = 120, k = 50.

我可以估计 n = 110, k = 50（返回 6x10^31）但不能运行 n = 120, k = 50。

回答by om prakash

using System;
//calculating factorial with recursion
namespace ConsoleApplication2
{
    class Program
    {
        long fun(long a)
        {
            if (a <= 1)
            {
                return 1;}
            else
            {
                long c = a * fun(a - 1);
                return c;
            }}

        static void Main(string[] args)
        {

            Console.WriteLine("enter the number");
            long num = Convert.ToInt64(Console.ReadLine());
            Console.WriteLine(new Program().fun(num));
            Console.ReadLine();
        }
    }
}

回答by arash jafarzadeh

The following can calculate the factorial of 5000 in 1 second.

以下可以在 1 秒内计算 5000 的阶乘。

public class Number
{
    #region Fields
    private static long _valueDivision = 1000000000;
    private static int _valueDivisionDigitCount = 9;
    private static string _formatZeros = "000000000";
    List<long> _value;
    #endregion

    #region Properties
    public int ValueCount { get { return _value.Count; } }
    public long ValueAsLong
    {
        get
        {
            return long.Parse(ToString());
        }
        set { SetValue(value.ToString()); }
    }
    #endregion

    #region Constructors
    public Number()
    {
        _value = new List<long>();
    }
    public Number(long value)
        : this()
    {
        SetValue(value.ToString());
    }
    public Number(string value)
        : this()
    {
        SetValue(value);
    }
    private Number(List<long> list)
    {
        _value = list;
    }
    #endregion

    #region Public Methods
    public void SetValue(string value)
    {
        _value.Clear();
        bool finished = false;
        while (!finished)
        {
            if (value.Length > _valueDivisionDigitCount)
            {
                _value.Add(long.Parse(value.Substring(value.Length - _valueDivisionDigitCount)));
                value = value.Remove(value.Length - _valueDivisionDigitCount, _valueDivisionDigitCount);
            }
            else
            {
                _value.Add(long.Parse(value));
                finished = true;
            }
        }
    }
    #endregion

    #region Static Methods
    public static Number operator +(Number c1, Number c2)
    {
        return Add(c1, c2);
    }
    public static Number operator *(Number c1, Number c2)
    {
        return Mul(c1, c2);
    }
    private static Number Add(Number value1, Number value2)
    {
        Number result = new Number();
        int count = Math.Max(value1._value.Count, value2._value.Count);
        long reminder = 0;
        long firstValue, secondValue;
        for (int i = 0; i < count; i++)
        {
            firstValue = 0;
            secondValue = 0;
            if (value1._value.Count > i)
            {
                firstValue = value1._value[i];
            }
            if (value2._value.Count > i)
            {
                secondValue = value2._value[i];
            }
            reminder += firstValue + secondValue;
            result._value.Add(reminder % _valueDivision);
            reminder /= _valueDivision;
        }
        while (reminder > 0)
        {
            result._value.Add(reminder % _valueDivision);
            reminder /= _valueDivision;
        }
        return result;
    }
    private static Number Mul(Number value1, Number value2)
    {
        List<List<long>> values = new List<List<long>>();
        for (int i = 0; i < value2._value.Count; i++)
        {
            values.Add(new List<long>());
            long lastremain = 0;
            for (int j = 0; j < value1._value.Count; j++)
            {
                values[i].Add(((value1._value[j] * value2._value[i] + lastremain) % _valueDivision));
                lastremain = ((value1._value[j] * value2._value[i] + lastremain) / _valueDivision);
                //result.Add(();
            }
            while (lastremain > 0)
            {
                values[i].Add((lastremain % _valueDivision));
                lastremain /= _valueDivision;
            }
        }
        List<long> result = new List<long>();
        for (int i = 0; i < values.Count; i++)
        {
            for (int j = 0; j < i; j++)
            {
                values[i].Insert(0, 0);
            }
        }
        int count = values.Select(list => list.Count).Max();
        int index = 0;
        long lastRemain = 0;
        while (count > 0)
        {
            for (int i = 0; i < values.Count; i++)
            {
                if (values[i].Count > index)
                    lastRemain += values[i][index];
            }
            result.Add((lastRemain % _valueDivision));
            lastRemain /= _valueDivision;
            count -= 1;
            index += 1;
        }
        while (lastRemain > 0)
        {
            result.Add((lastRemain % _valueDivision));
            lastRemain /= _valueDivision;
        }
        return new Number(result);
    }
    #endregion

    #region Overriden Methods Of Object
    public override string ToString()
    {
        string result = string.Empty;
        for (int i = 0; i < _value.Count; i++)
        {
            result = _value[i].ToString(_formatZeros) + result;
        }
        return result.TrimStart('0');
    }
    #endregion
}

class Program
{
    static void Main(string[] args)
    {
        Number number1 = new Number(5000);
        DateTime dateTime = DateTime.Now;
        string s = Factorial(number1).ToString();
        TimeSpan timeSpan = DateTime.Now - dateTime;
        long sum = s.Select(c => (long) (c - '0')).Sum();
    }
    static Number Factorial(Number value)
    {
        if( value.ValueCount==1 && value.ValueAsLong==2)
        {
            return value;
        }
        return Factorial(new Number(value.ValueAsLong - 1)) * value;
    }
}

回答by mehvan

hello everybody according to this solution i have my own solution where i calculate factorial of array 1D elements. the code is `int[] array = new int[5] { 4,3,4,3,8 };

大家好，根据这个解决方案，我有我自己的解决方案，我计算数组一维元素的阶乘。代码是 `int[] array = new int[5] { 4,3,4,3,8 };

        int fac = 1;

        int[] facs = new int[array.Length+1];

        for (int i = 0; i < array.Length; i++)
        {
            for (int j = array[i]; j > 0; j--)
            {
                fac *= j;
            }
            facs[i] = fac;
            textBox1.Text += facs[i].ToString() + " ";
            fac = 1;
        }`

copy and paste the code above ^ in the button , it solves factorial of elements of array 1D. best regards.

将上面 ^ 的代码复制并粘贴到按钮中，它解决了数组 1D 元素的阶乘。此致。