Quick & power matrix using matrix powers quickly find Fibonacci column item n

J cattle off first question made me understand myself how many things are still not today, to learn about God matrix (jian) odd (dan) and how to use the power of fast power matrix quickly find Fibonacci column n item (n> 1e10)

First on the code:

#include <iostream>
#include <cstring>
#include <string>
#define ll long long
#define mod 1000000007
using namespace std;
struct node {
    ll a[2][2];
    void init_1()                    //1 1
    {                                //0 0
        a[0][0] = a[0][1] = 1;
        a[1][0] = a[1][1] = 0;
    }

    void init_2()                     //1 1
    {                                 //1 0
        a[0][0] = a[0][1] = a[1][0] = 1;
        a[1][1] = 0;
    }

    void init_3()                     //1 0
    {                                 //0 1
        a[0][0] = a[1][1] = 1;
        a[0][1] = a[1][0] = 0;
    }

    friend ll getfabo(node a,node b,ll n)
    {
        if (n == 1)
            return 1;
        a.init_1();
        b.init_2();
        b = qmul(b,n-2);
        a = mul(a, b);
        return a.a[0][0];
    }

    friend node mul(node a,node b)//朴实无华的矩阵乘法
    {
        node ans;
        for (int i = 0; i < 2; i++)
        {
            for (int j = 0; j < 2; j++)
            {
                ans.a[i][j] = 0;
                for (int k = 0; k < 2; k++)
                {
                    ans.a[i][j] = (ans.a[i][j] + (a.a[i][k] * b.a[k][j] % mod)) % mod;
                }
            }
        }
        return ans;
    }

    friend node qmul(node a, ll n)//(神奇而简单的矩阵快速幂)和快速幂原理一样
    {
        node b;
        b.init_3();
        while (n)
        {
            if (n & 1)
                b = mul(a, b);
            n >>= 1;
            a = mul(a, a);
        }
        return b;
    }

}fabo1,fabo2;

int main()
{
    cout << getfabo(fabo1, fabo2, 1000000000000)<<endl;
    return 0;
}

A simple proof:

These are the power matrix fast & Fibonacci board of n items ...