C# - Finding Prime Factors of a Number
Given a number, we can find out what would be the prime factors of the number. But it is not easy to find out since we have to go through couple of iteratons.
Prime Factor For 9360 = 13 * 5 * 3 * 3 * 2 * 2 * 2 * 2
Note that all the numbers 13, 5, 3, 2 are prime numbers.
Source Code
using System;
using System.Collections.Generic;
using System.Text;
namespace SoftwareAndFinance
{
class Math
{
const int MAX_SIZE = 15000;
static bool IsPrimeNumber(int num)
{
bool bPrime = true;
int factor = num / 2;
int i = 0;
for (i = 2; i <= factor; i++)
{
if ((num % i) == 0)
bPrime = false;
}
return bPrime;
}
static public int GetPrimeFactors(int num, out int [] arrResult)
{
int count = 0;
int [] arr = new int[MAX_SIZE];
arrResult = new int[MAX_SIZE];
int i = 0;
int idx = 0;
for(i = 2; i <= num; i++)
{
if(IsPrimeNumber(i) == true)
arr[count++] = i;
}
while(true)
{
if(IsPrimeNumber(num) == true)
{
arrResult[idx++] = num;
break;
}
for(i = count - 1; i >= 0; i--)
{
if( (num % arr[i]) == 0)
{
arrResult[idx++] = arr[i];
num = num / arr[i];
break;
}
}
}
return idx;
}
static void Main(string[] args)
{
int [] arrResult;
Console.Write("Enter a number to find prime factor: ");
int n = Convert.ToInt32(Console.ReadLine());
int count = GetPrimeFactors(n, out arrResult);
for (int i = 0; i < count; i++)
{
Console.Write("{0,4:n}", arrResult[i]);
if (i != count - 1)
Console.Write(" * ");
}
Console.WriteLine();
}
}
}
Click here to get the C# project along with executable
Output
Enter a Number to find prime factor: 9360
Prime Factors For 9360 = 13 * 5 * 3 * 3 * 2 * 2 * 2 * 2
Press any key to continue . . .
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