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It 's ok.

I've just posted the Binet's formula because it's a curiosity, of course it slow down the PC , but if the user whats a number very big,maybe is it better use the formula? or ...

now it's compiling! :cool:




#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

simple, just include math.h , and than:

result = (1/sqrt(5))*((1+sqrt(5))/2)^n-(1/sqrt(5))*((1-sqrt(5))/2)^n ;

of course n is the variable...

i've learned this, this week in college. :):cool: