how to calculate 2^n modulo 1000000007 , n = 10^9

Question

what is the fastest method to calculate this, i saw some people using matrices and when i searched on the internet, they talked about eigen values and eigen vectors (no idea

Answer 1

Modular exponentiation by the square-and-multiply method:

function powerMod(b, e, m)
    x := 1
    while e > 0
        if e%2 == 1
            x, e := (x*b)%m, e-1
        else b, e := (b*b)%m, e//2
    return x

Answer 2

f(n) = (2*f(n-1)) + 2 with f(1)=1

is equivalent to

(f(n)+2) = 2 * (f(n-1)+2)
         = ...
         = 2^(n-1) * (f(1)+2) = 3 * 2^(n-1)

so that finally

f(n) = 3 * 2^(n-1) - 2

where you can then apply fast modular power methods.

Answer 3

C code for calculating 2^n

    const int mod = 1e9+7;

    //Here base is assumed to be 2
    int cal_pow(int x){
        int res;
        if (x == 0) res=1;
        else if (x == 1)    res=2;
        else {
            res = cal_pow(x/2);
            if (x % 2 == 0) 
                res = (res * res) % mod;
            else
                res = (((res*res) % mod) * 2) % mod;
        }
        return res;
    }