(a/b) mod n for large numbers?
sn = (f(n+2)+2*f(n)+f(n-1)-1)/3

n/d ≡ ((n % m)*k % m) (mod m)

k = (f*m + 1)/d
n*k = (n*(f*m + 1))/d = ((n*f)*m + n)/d = (n/d)*(f*m) + (n/d)

n*k ≡ n/d (mod m)

n*k ≡ (n % m)*k (mod m)

long mod_ans = 0;

for(i = 0; i < digits_array_length; i++)

{

    int digit = digits_array[i];
    mod_ans = (mod_ans * 16 + digit) % num;
}