pku 3070 Fibonacci 矩阵快速幂相乘求Fibonacci 数列

http://poj.org/problem?id=3070

思路：

这里n很大单纯的递推是O(N)会超时，所以要用矩阵快速幂优化；

f(n)            1    1                           f(1)

            =            的n-1次方 *

f(n-1)         1   0                          f(0)

 

View Code

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <set>

#include <map>

#include <string>



#define CL(a,num) memset((a),(num),sizeof(a))

#define iabs(x)  ((x) > 0 ? (x) : -(x))

#define Min(a,b) (a) > (b)? (b):(a)

#define Max(a,b) (a) > (b)? (a):(b)



#define ll __int64

#define inf 0x7f7f7f7f

#define MOD 10000

#define lc l,m,rt<<1

#define rc m + 1,r,rt<<1|1

#define pi acos(-1.0)

#define test puts("<------------------->")

#define maxn 10000007

#define M 100007

#define N 5

using namespace std;

//freopen("din.txt","r",stdin);

struct Mat{

    ll mat[3][3];

    void init(){

        for (int i = 0; i < 2; ++i){

            for (int j = 0; j < 2; ++j){

                mat[i][j] = 1;

            }

        }

        mat[1][1] = 0;

    }

};

Mat operator * (Mat a,Mat b){

    Mat c;

    int i,j,k;

    CL(c.mat,0);

    //矩阵乘积

    for (i = 0; i < 2; ++i){

        for (j = 0; j < 2; ++j){

            for (k = 0; k < 2; ++k){

                if (!a.mat[i][k] || !b.mat[k][j]) continue;

                c.mat[i][j] += a.mat[i][k]*b.mat[k][j];

                if (c.mat[i][j] > MOD) c.mat[i][j] %= MOD;

            }

        }

    }

    return c;

}

Mat operator ^ (Mat a,ll k){

    Mat c;

    int i,j;

    //单位矩阵

    for (i = 0; i < 2; ++i){

        for (j = 0; j < 2; ++j){

            c.mat[i][j] = (i == j);

        }

    }

    //求k次幂

    while (k){

        if (k&1) c = c*a;

        k >>= 1;

        a = a*a;

    }

    return c;

}



int main(){

    ll n;

    Mat a;

    while (~scanf("%I64d",&n)){

        if (n == -1) break;

        if (n == 0) {printf("0\n"); continue;}

        a.init();//初始化

        a = (a^(n - 1));//求n - 1次幂的到{fn,fn - 1};

        printf("%I64d\n",a.mat[0][0]);

    }

    return 0;

}