1. 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

2. 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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