c# - 我可以在 C# 中找到 BigInteger 的位数吗？

Question

我正在解决 this problem ，其中他们要求第一个 1000 位斐波那契数的索引，我的第一个想法类似于:

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，没有计算 BigInteger 位数的方法。这是真的？绕过它的一种方法是使用 BigInteger 的 .ToString().Length 方法，但有人告诉我字符串处理速度很慢。

BigInteger 也有一个 .ToByteArray()，我想将 BigInteger 转换为字节数组并检查该数组的长度 - 但我认为这不能唯一地确定 BigInteger 中的位数。

为了它的值(value)，我实现了另一种解决它的方法，即手动将斐波那契数存储在数组中，并在数组已满时立即停止，我将其与基于 .ToString 的方法进行了比较，后者慢了大约2.5倍，但是第一种方法用了0.1秒，也显得很长。

编辑:我已经测试了下面答案中的两个建议(一个使用 BigInteger.Log，另一个使用 MaxLimitMethod)。我得到以下运行时间:

• 原始方法:00:00:00.0961957
• 字符串方法:00:00:00.1535350
• BigIntegerLogMethod: 00:00:00.0387479
• 最大限制方法: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<n; i++)
            {
                singleSum = firstNumber[i] + secondNumber[i];
                if (singleSum >= 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;
    }
}

Answer 1

前提是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; 

方法。对于 100 000 000 次迭代，它比 Log10 解决方案慢 3 倍。