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Old 05-06-2008, 07:02 AM
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Default Iterative Algorithms
This is a sample of the iterative approach to Fibonacci numbers.

Code:
/**
 * @(#)FibonacciNumbers.java
 *
 *
 * @author 
 * @version 1.00 2008/5/5
 */


public class FibonacciNumbers 
{

    public FibonacciNumbers(long n) 
    {
    	System.out.println("Your number is: " + this.findNumbers(n));
    }
    
    private long findNumbers(long n)
    {
    	long temp1 = 1;
    	long temp2 = 0;
    	long fibonacciNumber = 0;
		for (long i = 2; i <= n; i++)
		{
			fibonacciNumber = temp1 + temp2;
			temp2 = temp1;
			temp1 = fibonacciNumber;
		}
		return fibonacciNumber;
    }
    
    public static void main(String[] args)
    {
    	FibonacciNumbers fibNum = new FibonacciNumbers(190);
    }
}
the number this will work for is 190
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Old 07-05-2008, 07:29 AM
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Hm, i think it is wrong. If n=190 the fibNum is more than 18 digit, while long int only 18 digit...

n = 190 , fibnum is more than 19 digit while long int has only 18 digit, what is the fibonacci of 100? The answer is 354224848179261915075 and it's more than 19 digit...

I think it is wrong. If(n==190) the answer is more than 19 digit while long only 19 digit...

CMIIW..

Last edited by CaptainMorgan; 08-01-2008 at 08:18 AM.
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