斐波那契数列的第N项

#include <bits/stdc++.h>
#define _xx ios_base::sync_with_stdio(0);cin.tie(0);
using namespace std;
typedef long long ll;
const ll p = 1000000009ll;
struct node
{
    ll data[2][2];
};
node operator * (const node& t1, const node& t2)
{
    node ans;
    for(int i = 0; i < 2; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            ans.data[i][j] = 0;
            for(int k = 0; k < 2; k++)
                ans.data[i][j] += t1.data[i][k]*t2.data[k][j];
                ans.data[i][j] %= p;
        }
    }
    return ans;
}
node operator % (const node& t1, const ll& p)
{
    node ans;
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
            ans.data[i][j] = t1.data[i][j]%p;
    return ans;
}
node fastm(node a, ll b)
{
    node ans;
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
            if(i == j) ans.data[i][j] = 1;
            else ans.data[i][j] = 0;
    while(b)
    {
        if(b%2 == 1)
        {
            ans = ans*a;
            ans = ans%p;
        }
        a = a*a;
        a = a%p;
        b /= 2;
    }
    return ans;
}
int main()
{_xx
    ll n;
    cin >> n;
    if(n == 0) cout << 0 << endl;
    else if(n == 1) cout << 1 << endl;
    else
    {
        node a, b;
        for(int i = 0; i < 2; i++)
            for(int j = 0; j < 2; j++)
                b.data[i][j] = a.data[i][j] = 0;
        a.data[1][0] = a.data[2][0] = 1;
        b.data[0][1] = b.data[1][0] = b.data[1][1] = 1;
        b = fastm(b, n - 1);
        a = b*a;
        cout << a.data[1][0]%p << endl;
    }
    return 0;
}