This is a sample of the iterative approach to Fibonacci numbers.
the number this will work for is 190
* @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;
public static void main(String args)
FibonacciNumbers fibNum = new FibonacciNumbers(190);
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...