矩阵乘法加速fib数列

考虑矩阵（1,1）（1,0）

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
#define LL long long
const int mod = 1e9+;
const int N = ;
struct Matrix{
    int a[N][N];
    Matrix(){
        this -> clear();
    }
    void clear(){
        memset(a,,sizeof(a));
    }
    void setone(){
        this ->clear();for(int i=;i<N;++i)
            a[i][i]=;
    }

    Matrix operator * (const Matrix &x) {
         Matrix c;
         for (int k=;k<N;++k)
             for (int i=;i<N;++i)
                 for (int j=;j<N;++j)
                    c.a[i][j]=(c.a[i][j]+(LL)a[i][k]*x.a[k][j])%mod;
        return c;
    }
};
int fibn(int x)
{
    Matrix a,b;
    a.clear();
    a.a[][]=a.a[][]=a.a[][]=;
    b.setone();
    while(x)
    {
        if(x&) b=b*a;
        a=a*a;
        x>>=;
    }
    return (b.a[][]+b.a[][])%mod;
}
int main()
{
    int n;
    while (~scanf("%d",&n))
    {
        if (n==||n==) puts("");
        else printf("%d\n",fibn(n-));
    }
    return ;
}