sum of even fibonacci numbers under 4000000

gistfile1.c

#include <stdio.h>
#include <math.h>

int main(void) {

  int N = 4000000;
  double ratio = 1.61803399;
  int n;
  int sum = 0;
  double tmp = 0.0;
  double fib_n = 0.0;

  for (n = 3; fib_n < N; n += 3) {
    if (n == 3) {
      fib_n = 2;
    } else {
      tmp = round(fib_n * ratio);
      fib_n = tmp + round(tmp * ratio);
    }
    printf("fib_%d = %.0f\n", n, fib_n);
    if (fib_n < N) {
      sum += fib_n;
    }
  }

  printf("sum = %d\n", sum);

  return 0;
}

If you're using pre-C99 version of C, usefloor() instead of round():
tmp = floor(fib_n * ratio + 0.5)
fib_n = tmp + floor(tmp * ratio + 0.5);