poj 3070 Fibonacci (矩阵的快速幂)

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

 

题意：

已知  

Fibonacci 数列 可以这样求 ，求 fn 的最后四位 

.

 题解：

  矩阵的快速幂

 

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<string>
#define Min(a,b) a<b?a:b
#define Max(a,b) a>b?a:b
#define CL(a,num) memset(a,num,sizeof(a));
#define maxn  40
#define eps  1e-6
#define inf 9999999
#define mx 1<<60
#define mod 10000
using namespace std;
struct martrix
{
    int m[3][3];
};
martrix mtmul(martrix a,martrix b)
{
    martrix c;
    int i,j,k;
    for(i = 0; i < 2; i++)
    {
        for(j = 0; j < 2;j++)
        {
            c.m[i][j] = 0;
            for(k = 0 ; k < 2;k++)
            {
                c.m[i][j] += a.m[i][k] * b.m[k][j];
                c.m[i][j] %=mod;
            }
        }
    }


    return c;
}
martrix mtpow(martrix d,int k)
{   martrix a;
    if(k == 1) return d;
    int mid = k / 2;
    a = mtpow(d,k/2);
    a = mtmul(a,a);
    if(k & 1)
    {
        a = mtmul(a,d);
    }
    return a;


}
int main()
{
    __int64 n;
    martrix a,b;
    a.m[0][0] = 1;
    a.m[0][1] = 1;
    a.m[1][0] = 1;
    a.m[1][1] = 0 ;
    while(scanf("%I64d",&n),n >= 0)
    {
        if(n == 0)printf("0\n");
        else{

          b = mtpow(a,n);
          printf("%d\n",b.m[1][0] % mod);

        }
    }
}