my idea :
typical sequence

i      1 2 3 4 5 6 7
fib(i) 1 1 2 3 5 8 13

assume that i<=j
first, we have to find f(i+1)

diff=j-i
using F(i+1)=Fj/fib(diff) - Fi*fib(diff-1)/fib(diff)

but my prog didn't compute fib(i) but only computed
1/fib(i)  and  fib(i-1)/fib(i)

using formular
1/fib(i) = 1/(1/fib(i-1)+1/fib(i-2))
double f1(int ind)// return 1/fib(ind)
{
    int i;
    double a[3]={1,1};
    for(i=2;i<ind;i++)
        a[i%3]=1/(1/a[(i+1)%3]+1/a[(i+2)%3]);
    return a[(ind -1)%3];
}

fib(i-1)/fib(i) using this relation
1+1/(1+1/(1+1/...))
double f2(int ind)// return fib(ind-1)/fib(ind)
{
    int i;
    double ret=1;
    for(i=0;i<ind-2;i++)
        ret=1/(ret+1);
    return ret;
}

using this procedure, we can store any computation with
double and avoid int overflow.

and so the F(i+1) = (Fj*f1(diff)-Fi*f2(diff)), then find f(n) with ordinary algo using int(it said that fk can store in INT) !!