Question

I’m solving this problem:

G(n) is defined as

G(n) = G(n-1) + f(4n-1) , for n > 0

and G(0) = 0

f(i) is ith Fibonacci number. Given n you need to evaluate G(n)

modulo 1000000007.

Input

First line contains number of test cases t (t<40000). Each of the next t

lines contain an integer n ( 0 <= n <
2^51).

Output

For each test case print G(n) modulo 1000000007.

Example

Input:
2
2
4



Output:


15
714

This is the code I’ve written:

typedef long long type;
#define check 1000000007
type x;
type y;

type f(type n)
{
    return(ceil((pow(1.618,n) - pow(-0.618,n))/((sqrt(5)*1.0))));
}
type val(type n)
{
    if(n==0)
    return 0;
    else 
    return (val(n-1)+f(4*n-1));
}
int main()
{
    cin>>x;
    while(x--)
    {
       cin>>y;
       cout<<val(y)%check<<endl;
       }
    //getch();
    return 0;
}

Can you suggest any improvements?

Answer 1

Sometimes such problems can be tackled with mathematical tricks,
instead of brute force solutions.

The large value of n and modulo, in my opinion, are indications that
a clever solution exists. Of course figuring out the solution is the hard part.

(I’m not sure if this is ideal in your case, I’m only pointing you an alternative way)

For example, in the Art of Computer Programming, Volume 1: Fundamental Algorithms
Knuth uses “generating functions”, a clever way for constructing a closed form
for the Fn fibonacci number.

For more info read Generating Functions (pdf)

Why should one care about the
generating function for a sequence?
There are several answers, but here is
one: if we can find a generating
function for a sequence, then we can
often find a closed form for the nth
coefficient— which can be pretty
useful! For example, a closed form for
the coefficient of xⁿ in the
power series for x/(1−x−x²)
would be an explicit formula for the
nth Fibonacci number. […]