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)

 

#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;
}