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  1. #1
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    Zosden is offline Senior Member
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    Default Iterative Algorithms

    This is a sample of the iterative approach to Fibonacci numbers.

    Java 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
    My IP address is 127.0.0.1

  2. #2
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    Default

    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 07:18 AM.

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