我可以在C#中找到BigInteger的位数吗？

我正在解决这个问题,他们要求第一个斐波纳契数为1000位的索引,我的第一个想法是类似于:

BigInteger x = 1;
BigInteger y = 1;
BigInteger tmp = 0;

int currentIndex = 2;
while (x.NoOfDigits < 1000)
{
    tmp = x + y;
    y = x;
    x = tmp;
    currentIndex++;
}
return currentIndex;

但是,据我所知,没有方法可以计算BigInteger的位数.这是真的？绕过它的一种方法是使用BigInteger的.ToString().Length方法,但我被告知字符串处理很慢.

BigInteger也有一个.ToByteArray(),我想把BigInteger转换成一个字节数组并检查该数组的长度 - 但我不认为这唯一地决定了BigInteger中的位数.

为了它的价值,我实现了另一种解决方法,即手动将Fibonacci数存储在数组中,并在数组填满后立即停止,并将其与基于.ToString的方法进行比较,大约为2.5时间慢,但第一种方法需要0.1秒,这也似乎很长一段时间.

编辑:我已经在下面的答案中测试了这两个建议(一个是BigInteger.Log,一个是MaxLimitMethod).我得到以下运行时间:

原始方法:00:00:00.0961957

StringMethod:00:00:00.1535350

BigIntegerLogMethod:00:00:00.0387479

MaxLimitMethod:00:00:00.0019509

程序

using System;
using System.Collections.Generic;
using System.Numerics;
using System.Diagnostics;

class Program
{
    static void Main(string[] args)
    {
        Stopwatch clock = new Stopwatch();
        clock.Start();
        int index1 = Algorithms.IndexOfNDigits(1000);
        clock.Stop();
        var elapsedTime1 = clock.Elapsed;
        Console.WriteLine(index1);
        Console.WriteLine("Original method: {0}",elapsedTime1);
        Console.ReadKey();

        clock.Reset();
        clock.Start();
        int index2 = Algorithms.StringMethod(1000);
        clock.Stop();
        var elapsedTime2 = clock.Elapsed;
        Console.WriteLine(index2);
        Console.WriteLine("StringMethod: {0}", elapsedTime2);
        Console.ReadKey();

        clock.Reset();
        clock.Start();
        int index3 = Algorithms.BigIntegerLogMethod(1000);
        clock.Stop();
        var elapsedTime3 = clock.Elapsed;
        Console.WriteLine(index3);
        Console.WriteLine("BigIntegerLogMethod: {0}", elapsedTime3);
        Console.ReadKey();

        clock.Reset();
        clock.Start();
        int index4 = Algorithms.MaxLimitMethod(1000);
        clock.Stop();
        var elapsedTime4 = clock.Elapsed;
        Console.WriteLine(index4);
        Console.WriteLine("MaxLimitMethod: {0}", elapsedTime4);
        Console.ReadKey();


    }
}

static class Algorithms
{
    //Find the index of the first Fibonacci number of n digits
    public static int IndexOfNDigits(int n)
    {
        if (n == 1) return 1;
        int[] firstNumber = new int[n];
        int[] secondNumber = new int[n];

        firstNumber[0] = 1;
        secondNumber[0] = 1;
        int currentIndex = 2;

        while (firstNumber[n-1] == 0)
        {
            int carry = 0, singleSum = 0;
            int[] tmp = new int[n]; //Placeholder for the sum
            for (int i = 0; i= 10) carry = 1;
                else carry = 0;

                tmp[i] += singleSum % 10;
                if (tmp[i] >= 10)
                {
                    tmp[i] = 0;
                    carry = 1;
                }
                int countCarries = 0;
                while (carry == 1)
                {
                    countCarries++;
                    if (tmp[i + countCarries] == 9)
                    {
                        tmp[i + countCarries] = 0;
                        tmp[i + countCarries + 1] += 1;
                        carry = 1;
                    }
                    else
                    {
                        tmp[i + countCarries] += 1;
                        carry = 0;
                    }
                }
            }

            for (int i = 0; i < n; i++ )
            {
                secondNumber[i] = firstNumber[i];
                firstNumber[i] = tmp[i];
            }
            currentIndex++;
        }
        return currentIndex;
    }

    public static int StringMethod(int n)
    {
        BigInteger x = 1;
        BigInteger y = 1;
        BigInteger tmp = 0;
        int currentIndex = 2;

        while (x.ToString().Length < n)
        {
            tmp = x + y;
            y = x;
            x = tmp;
            currentIndex++;
        }
        return currentIndex;
    }

    public static int BigIntegerLogMethod(int n)
    {
        BigInteger x = 1;
        BigInteger y = 1;
        BigInteger tmp = 0;
        int currentIndex = 2;

        while (Math.Floor(BigInteger.Log10(x) + 1) < n)
        {
            tmp = x + y;
            y = x;
            x = tmp;
            currentIndex++;
        }
        return currentIndex;
    }

    public static int MaxLimitMethod(int n)
    {
        BigInteger maxLimit = BigInteger.Pow(10, n - 1);
        BigInteger x = 1;
        BigInteger y = 1;
        BigInteger tmp = 0;
        int currentIndex = 2;

        while (x.CompareTo(maxLimit) < 0)
        {
            tmp = x + y;
            y = x;
            x = tmp;
            currentIndex++;
        }
        return currentIndex;
    }
}

PHeiberg.. 8

假设x> 0

int digits = (int)Math.Floor(BigInteger.Log10(x) + 1);

会得到位数.

出于好奇,我测试了

int digits = x.ToString().Length; 

做法.对于10万次迭代,它比Log10解决方案慢3倍.

假设x> 0

int digits = (int)Math.Floor(BigInteger.Log10(x) + 1);

会得到位数.

出于好奇,我测试了

int digits = x.ToString().Length; 

做法.对于10万次迭代,它比Log10解决方案慢3倍.

扩展我的评论 - 而不是基于数字的测试,基于超过具有问题上限的常量进行测试:

public static int MaxLimitMethod(int n)
    {
        BigInteger maxLimit = BigInteger.Pow(10, n);
        BigInteger x = 1;
        BigInteger y = 1;
        BigInteger tmp = 0;
        int currentIndex = 2;

        while (x.CompareTo(maxLimit) < 0)
        {
            tmp = x + y;
            y = x;
            x = tmp;
            currentIndex++;
        }
        return currentIndex;
    }

这应该会导致显着的性能提升.