Fibonacci--poj3070(矩阵快速幂)

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

 

 

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.

 

 

 

这道题就是快速幂http://blog.csdn.net/u013795055/article/details/38599321

还有矩阵相乘   

 

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#include <vector>
#include <queue>

using namespace std;
#define memset(a,b) memset(a,b,sizeof(a))
#define N 4
#define INF 0xfffffff
struct node
{
    int a[N][N];
}e;

node mm(node p,node q)
{
    node t;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            t.a[i][j]=0;
            for(int k=0;k<2;k++)
            {
                t.a[i][j]=(t.a[i][j]+(p.a[i][k]*q.a[k][j]))%10000;
            }
        }
    }
    return t;
}


node mul(node p,int n)
{

    node q;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            q.a[i][j]=(i==j);
        }
    }///这个是为了当n是奇数是第一次跟q相乘是还是原来的p,为了下一次跟下一奇数相乘

    while(n)
    {
        if(n&1)
           q=mm(q,p);
        n=n/2;
        p=mm(p,p);
    }
    return q;
}
int main()
{
    int n;
    while(scanf("%d",&n),n!=-1)
    {
        e.a[0][0]=1;
        e.a[0][1]=1;
        e.a[1][0]=1;
        e.a[1][1]=0;
        node b=mul(e,n);
        printf("%d\n",b.a[0][1]);
    }
    return 0;
}