5

-----请求修改代码--------

问题：数一下有 50 个因数的快速三角形系列数？

详细说明：假设有一个系列

   1 : 1
   3 : 1+2
   6 : 1+2+3 
   10 : 1+2+3+4
   15 : 1+2+3+4+5
   21 : 1+2+3+4+5+6
   28 : 1+2+3+4+5+6+7

这里 1,3,6,10,15,21,28 是三角形系列下的数字。

让我们看看数字的因素

    Number factors         Count
    1     : 1               1              
    3     : 1,3             2
    6     : 1,2,3,6         4
    10    : 1,2,5,10        4
    15    : 1,3,5,15        4
    21    : 1,3,7,21        4
    28    : 1,2,4,7,14,28   6

这里 6 是第一个具有 4 个因子的三角形数。即使 10、15、21 也有 4 个因子，但它们不是第一个。像这样让我们把一个数字作为 2，它有 2 个因子作为 1 和 2，对于数字 3 也有 2 个因子作为 1 和 3

但根据问题 3 将不是 2 的答案，因为 2 不在三角形序列号列表中，即使它比 3 快。

4

4 回答 4

5

三角形数 #2591 = 3357936 是第一个正好有50 个因子的三角形数：1、2、3、4、6、8、9、12、16、18、24、27、36、48、54、72、81， 108，144，162，216，324，432，648，1296，2591，5182，7773，10364，15546，20728，23319，31092，41456，41456，466638，466638，62184，62184，699957，699957，93276，193276，2044,104,104,104,104,104,104,104,134,104,104,104,104,104,104,104,104,104,134,104,104,134,104,104,104,104,104,104,104,10 ,， 279828、373104、419742、559656、839484、1119312、1678968、3357936

三角形数 #12375 = 76576500 是第一个具有至少500 个因子（实际上是 576 个因子）的三角形：1、2、3、4、5、6、7、9、10、11、...、19144125、25525500、 38288250, 76576500

三角形数 #1569375 = 1231469730000 是第一个正好有500 个因子的

解决方案代码本身非常简单，只要您可以获得除数：

   public static long Solution(int factorsCount) {
      for (long i = 1; ; ++i) {
        long n = i * (i + 1) / 2;

        IList<long> factors = GetDivisors(n);

        // This code tests if a triangle number has exactly factorsCount factors
        // if you want to find out a triangle number which has at least factorsCount factors
        // change "==" comparison to ">=" one:
        // if (factors.Count >= factorsCount)  
        if (factors.Count == factorsCount) 
          return n;
      }
    }

  ...

  long solution = Solution(50);

如果您没有例程来获取数字的因素，您可以使用这个：

// Get prime divisors 
private static IList<long> CoreGetPrimeDivisors(long value, IList<int> primes) {
  List<long> results = new List<long>();

  int v = 0;
  long threshould = (long) (Math.Sqrt(value) + 1);

  for (int i = 0; i < primes.Count; ++i) {
    v = primes[i];

    if (v > threshould)
      break;

    if ((value % v) != 0)
      continue;

    while ((value % v) == 0) {
      value = value / v;

      results.Add(v);
    }

    threshould = (long) (Math.Sqrt(value) + 1);
  }

  if (value > 1)
    results.Add(value);

  return results;
}

/// <summary>
/// Get prime divisors 
/// </summary>
public static IList<long> GetPrimeDivisors(long value, IList<int> primes) {
  if (!Object.ReferenceEquals(null, primes))
    return CoreGetPrimeDivisors(value, primes);

  List<long> results = new List<long>();

  while ((value % 2) == 0) {
    results.Add(2);

    value = value / 2;
  }

  while ((value % 3) == 0) {
    results.Add(3);

    value = value / 3;
  }

  while ((value % 5) == 0) {
    results.Add(5);

    value = value / 5;
  }

  while ((value % 7) == 0) {
    results.Add(7);

    value = value / 7;
  }

  int v = 0;
  long n = (long) (Math.Sqrt(value) / 6.0 + 1);
  long threshould = (long) (Math.Sqrt(value) + 1);

  for (int i = 2; i <= n; ++i) {
    v = 6 * i - 1;

    if ((value % v) == 0) {
      while ((value % v) == 0) {
        results.Add(v);

        value = value / v;
      }

      threshould = (long) (Math.Sqrt(value) + 1);
    }

    v = 6 * i + 1;

    if ((value % v) == 0) {
      while ((value % v) == 0) {
        results.Add(v);

        value = value / v;
      }

      threshould = (long) (Math.Sqrt(value) + 1);
    }

    if (v > threshould)
      break;
  }

  if (value > 1) {
    if (results.Count <= 0)
      results.Add(value);
    else if (value != results[results.Count - 1])
      results.Add(value);
  }

  return results;
}

/// <summary>
/// Get all divisors
/// </summary>
public static IList<long> GetDivisors(long value, IList<int> primes) {
  HashSet<long> hs = new HashSet<long>();

  IList<long> divisors = GetPrimeDivisors(value, primes);

  ulong n = (ulong) 1;
  n = n << divisors.Count;

  for (ulong i = 1; i < n; ++i) {
    ulong v = i;
    long p = 1;

    for (int j = 0; j < divisors.Count; ++j) {
      if ((v % 2) != 0)
        p *= divisors[j];

      v = v / 2;
    }

    hs.Add(p);
  }

  List<long> result = new List<long>();

  result.Add(1);

  var en = hs.GetEnumerator();

  while (en.MoveNext())
    result.Add(en.Current);

  result.Sort();

  return result;
}

/// <summary>
/// Get all divisors
/// </summary>
public static IList<long> GetDivisors(long value) {
  return GetDivisors(value, null);
}
于 2013-07-09T06:48:19.720 回答
3

这是我的答案

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace TriangularSeries
{
    class MyClass
    {
        static void Main(string[] args)
        {
            int result;
            TriangularSeries aSeries = new TriangularSeries();
            result = aSeries.TSeries();
            Console.WriteLine("The first Triangular Series number that has 50Factors is : " + result);
            Console.Read();
        }
    }

    //Find the Triangular Series numbers
    class TriangularSeries
    {
        public int TSeries()
        {
            int fCount = 0, T1 = 1, i = 1, T2 = 0, fval = 0;
            while (fCount != 50)
            {
                i += 1;
                T2 = T1 + i;

                fCount = CalcFactors(T1);
                fval = T1;                   
                T1 = T2;

            }
            return fval;
        }

        public int CalcFactors(int num1)
        {

            List<int> factors = new List<int>();
            int max = (int)Math.Sqrt(num1);  //round down
            for (int factor = 1; factor <= max; ++factor)
            {
                //test from 1 to the square root, or the int below it, inclusive.
                if (num1 % factor == 0)
                {
                    factors.Add(factor);
                    if (factor != num1 / factor)
                    {
                        // Don't add the square root twice!  
                        factors.Add(num1 / factor);
                    }
                }
            }
            return factors.Count;

        }
    }   
}
于 2013-11-30T09:36:28.737 回答
3

解决方案：让我将问题分解为多个模块。

1）找到三角形系列直到一个数字。

2) 将所有标识的数字存储在整数列表中

3）找出特定数字的因子数

4）循环遍历三角形系列的每一项，并找到每个数字的因子数。

5）检查第一个计数为50的，然后显示该值

6）写break语句只显示前50个数字。

程序

using System;
using System.Collections.Generic;
using System.Collections;
using System.Linq;
using System.Text;

namespace IsNumberTringularSeriesConsoleApp
{ 
    class Program
    {
        /// <summary>
        /// Listing all numbers comes under Triangular series.
        /// </summary>
        /// <param name="number"></param>
        /// <returns></returns>
        static List<int> GetTriangularNumbers(int number)
        {
            List<int> lstTriangularNumbers = new List<int>();
            int i;
            int sum = 0;
            int triangularNumber = 0;
            for (i = 1; i < number; i++)
            {
                sum = sum + i;
                triangularNumber = sum;
                lstTriangularNumbers.Add(triangularNumber);
            }
            return lstTriangularNumbers;
        }

        /// <summary>
        /// returns(count) the number of factors for each number
        /// </summary>
        /// <param name="number"></param>
        /// <returns></returns>
        public static int FactorCount(int number)
        {
            List<int> factors = new List<int>();
            int max = (int)Math.Sqrt(number);  //round down
            for (int factor = 1; factor <= max; ++factor)
            { 
                //test from 1 to the square root, or the int below it, inclusive.
                if (number % factor == 0)
                {
                    factors.Add(factor);
                    if (factor != number / factor)
                   {
                     // Don't add the square root twice!  
                        factors.Add(number / factor);
                   }
                }
            }
            return factors.Count;
        }

        static void Main(string[] args)
        {
            List<int> lstTriangularNumbers = new List<int>();
            List<int> factors = new List<int>();
            int count = 0;
            //Getting the list of numbers comes under triangular series till 5000
            lstTriangularNumbers = GetTriangularNumbers(5000);

            foreach (int number in lstTriangularNumbers)
            {
                /*
                 * Calling the FactorCount(number) function to check no of factors 
                 * available for the specific triangular number - number.
                 */
                 count = FactorCount(number);
                 //Console.WriteLine("No of factors for : " + number + " is : " + count);
                if (count == 50)
                {
                    Console.WriteLine("No of factors for first Triangular Number : " + number + " is : " + count);
                    break;
                }
            }
            Console.ReadLine();
        }
    }
}
于 2013-07-09T06:30:17.997 回答
1

这是我的 C 语言程序

#include<stdio.h>
int i;
int num1=0,num2=1;
int a[3000];
int tri_series()         //This function finds the Triangular series numbers
{
    for(i=0;num2<=3000;i++)
    {
    num1=num1+num2;
    a[i]=num1;
    num2++;
    }
}
int main()
{
tri_series();            //Calling the function tri_series
int num,count;
    for(i=0;i<=3000;i++)
    {
      count=0;
      for(num=1;num<=a[i];num++)
      {
        if(a[i]%num==0)  //Finds the factors of each Triangular Series Number 
        count=count+1;   
      }
      if(count==50)      //Break at the first Triangular Series Number having 50 factors
      {
       printf("%d:%d\t",a[i],count);
       break;
      }
    }
}

性能问题：此代码在执行时间方面会产生性能问题。执行和显示输出需要“一分钟”。

于 2013-12-12T17:22:00.400 回答