nth term of series

Question

we have to find the nth term of this series http://oeis.org/A028859

n<=1000000000

answer should be modulo 1000000007

i have written the code but time limit exceeds when n a is huge number.

#include<iostream>
using namespace std

int main()
{
    long long int n;
    cin>>n;

    long long int a,b,c;
    a=1;
    b=3;

    int i;
    for(i=3;i<=n;i++)
    {
        c=(2ll*(a+b))%1000000007;
        a=b;
        b=c; 
    }

    cout<<c;
}

Answer 1

The standard technique for solving this type of problem is to rewrite it as a matrix multiplication and then use exponentiation by squaring to efficiently compute powers of the matrix.

In this case:

a(n+2) = 2 a(n+1) + 2 a(n)
a(n+1) = a(n+1)

(a(n+2)) = (2  2) * ( a(n+1) )
(a(n+1))   (1  0)   ( a(n)   )

So if we define the matrix A=[2,2 ; 1,0], then you can compute the n th term by

[1,0] * A^(n-2) * [3;1]

All of these operations can be done modulo 1000000007 so no big number library is required.

It requires O(log(n)) 2*2 matrix multiplications to compute A^N, so overall this method is O(log(n)), while your original method was O(n).

EDIT

Here is a good explanation and a C++ implementation of this method.

Answer 2

When long long is not enough, you probably want to use a bignum library. For example GNU MP.