-----修改所要求的代码--------
问题：计算具有50个因子的快速三角系列号？
详细说明：假设有一系列
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个因子的三角数。 即使10,15,21也有4个因素，但它们不是第1个。 就像那样，让一个数字为2，其中有2个因子为1和2 对于数字3也相同，也有2个因子为1和3
但是根据问题3，答案不是2，因为即使它快于3，2也不会进入三角系列号列表。
答案 0 :(得分：5)
三角数＃2591 = 3357936是第一个完全 50个因子： 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， 2591,5182,7773,10364,15546,20728,23319,31092,4145,46638,62184,69957,93276,124368,139914,186552,209871,279828,373104,419742,559656,839484,1199312,1678968,3357936 < / p>
三角数＃12375 = 76576500是第一个至少 500个因子（实际上是576个因子）：1,2,3,4,5,6,7,9,10， 11，...，19144125,25525500,38288250,76576500
三角形编号＃1569375 = 1231469730000是第一个完全 500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);
如果你还没有获得数字因素的例行程序，你可以使用这个：
// 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);
}
答案 1 :(得分：3)
解决方案： 让我在多个模块中分解问题。
1）找到三角形系列直到数字。
2）将所有已识别的数字存储在整数列表中
3）找出特定数字的因子数
4）循环通过每个三角形系列项目，找到每个数字的因子计数。
5）检查计数为50的第一个，然后显示值
6）写break语句只显示第50个数字。
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();
}
}
}
答案 2 :(得分：2)
这是我的回答
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;
}
}
}
答案 3 :(得分：1)
这是我用C语言编写的程序
#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： 此代码在执行时产生性能问题。执行并显示输出需要“一分钟”。