sicily Fibonacci 2

矩阵快速幂，1001. Fibonacci 2求斐波那契第n项！毕竟数据量太大！

http://soj.sysu.edu.cn/show_problem.php?pid=1001&cid=1740

#include <iostream>
#include <cstring>

using namespace std;

struct mat {
    long long m[2][2];
    mat()
    {
        memset(m, 0, sizeof(m));
    }
};

int n=2;

mat operator * (mat a, mat b)
{
    mat c;
    for(int i=0; i<n; i++)
        for(int j=0; j<n; j++)
            for(int k=0; k<n; k++)
                c.m[i][j] += (a.m[i][k] * b.m[k][j]) % 1000000007; //记得模 
    return c;
}

void print(mat a)
{
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
            cout << a.m[i][j] << " ";
        cout << endl;
    }
}

int main()
{
    int t;
    while(cin >> t)
    {
        if(t == -1)
            break;
        mat a;
        mat c;
        c.m[0][0] = 1; c.m[0][1] = 0; c.m[1][0] = 0; c.m[1][1] = 1;
        a.m[0][0] = 1; a.m[0][1] = 1; a.m[1][0] = 1; a.m[1][1] = 0;
        int N = t;
        while(N)
        {
            if(N%2) 
                c = c * a;
            a = a * a;
            N /= 2;
        }
        cout << c.m[0][1]%1000000007 << endl;
    }
    
    return 0;
}