斐波那契数列的第N项

斐波那契数列的第N项 51Nod – 1242

斐波那契数列的定义如下:

F(0) = 0
F(1) = 1
F(n) = F(n - 1) + F(n - 2) (n >= 2)

(1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, …)
给出n,求F(n),由于结果很大,输出F(n) % 1000000009的结果即可。
Input
输入1个数 n(1<=n<=1018) n ( 1 <= n <= 10 1 8 )
Output
输出 F(n)%1000000009 F ( n ) % 1000000009 的结果。
Sample Input

11

Sample Output

89

我的代码

#include <iostream>

using namespace std;
const long long int mod = 1000000009;

struct matrix
{
    long long int m[2][2];
    matrix()//赋初始值为0
    {
        for(int i = 0; i < 2; i++)
        {
            for(int j = 0; j < 2; j++)
            {
                m[i][j] = 0;
            }
        }
    }
};

matrix mul(matrix A, matrix B)//矩阵相乘
{
    matrix res;
    for(int i = 0; i <2; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            res.m[i][j] = 0;
            for(int k = 0; k < 2; k++)
            {
                res.m[i][j] += A.m[i][k]*B.m[k][j]%mod;
            }
        }
    }
    return res;
}

matrix pow(matrix A, long long int n)//矩阵求幂
{
    matrix imatrix;//单位矩阵
    imatrix.m[0][0] = 1;
    imatrix.m[1][1] = 1;
    if(n == 0)
    {
        return imatrix;
    }
    if(n%2 == 0)
    {
        return pow(mul(A,A),n/2);
    }
    else
    {
        return mul(pow(mul(A,A),n/2),A);
    }
}

int main()
{
    matrix A,ans;
    A.m[0][0] = 1;
    A.m[0][1] = 1;
    A.m[1][0] = 1;
    ans.m[0][0] = 1;
    long long int n;
    while(cin>>n)
    {
        matrix B = mul(ans,pow(A,n-1));
        cout << B.m[0][0] << endl;
    }
    return 0;
}
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