Question

#include <stdlib.h>
#include <stdio.h>
int main (){
   int n, cont, fib, na = 0, nb = 1, sum_even = 0;
   printf ("Insert a number and I'll tell you the respective Fibonacci: ");
   scanf ("%d", &n);
   for (cont = 1; cont < n; cont++) {
      na += nb;
      nb = na - nb;
      fib = na + nb;
      if (fib % 2 == 0) {
         sum_even += fib;
      }
   }
   printf ("%d\n", sum_even);
   return 0;
}

I was trying to do the Project Euler Problem 2, and then I came up with this code. The problem is: I can’t find the sum of the pair numbers on fibonacci’s sequence for numbers over 400 or something near that, because memory overflows. In consequence, I cant solve the exercise, since it asks to find the sum of the pair numbers below 4000000 in fibonacci’s sequence. Can anyone help me?

Edit:
I tried to used float type numbers to increase the answer’s capacity, it seems to work till a thousand or so, but if I try with bigger numbers, I get a -nan error in bash after like 15 secs of processing (I don’t really know what it means).

#include <stdlib.h>
#include <stdio.h>
int main () {
   int n, cont, div;
   float sum_even = 0, na = 0, nb = 1, fib;
   printf ("Insert a number and I'll tell you the respective Fibonacci: ");
   scanf ("%d", &n);
   for (cont = 1; cont <= n; cont++) {
      na += nb;
      nb = na - nb;
      fib = na + nb;
      div = fib / 2;
      if (div % 2 == 0) {
         sum_even += fib;
      }
   }
   printf ("%f\n", sum_even);
   return 0;
}

Answer 1

You misunderstood the problem statement. The task is to find the sum of

{ fib(n) : fib(n) <= 4000000 && fib(n) % 2 == 0 }

and not

{ fib(n) : n <= 4000000 && fib(n) % 2 == 0 }

That task is solved without problems with a minor modification to your code. Instead of

for (cont = 1; cont < n; cont++) {

use

while(fib <= n) {