poj 3070 矩阵快速幂

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12457		Accepted: 8851

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:

.

Source

Stanford Local 2006

 

题意： 求斐波那契数列

 

题解； 矩阵快速幂 第一题

        关键在于矩阵的构造  之后就是快速幂了

        矩阵的构造 http://www.cnblogs.com/frog112111/archive/2013/05/19/3087648.html

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<map>
#define mod 10000
using namespace std;
struct matrix
{
    int m[5][5];
} ans,exm;

struct matrix matrix_mulit(struct matrix aa,struct matrix bb)
{
    struct matrix there;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            there.m[i][j]=0;
            for(int k=0;k<2;k++)
            there.m[i][j]=(there.m[i][j]+aa.m[i][k]*bb.m[k][j]%mod)%mod;
        }
    }
    return there;
}
int matrix_quick(int gg)
{
     exm.m[0][0]=exm.m[0][1]=exm.m[1][0]=1;
     exm.m[1][1]=0;
     ans.m[0][0]=ans.m[1][1]=1;  
     ans.m[0][1]=ans.m[1][0]= 0;
     while(gg)
     {
         if(gg&1)  
         {
             ans=matrix_mulit(ans,exm);        
         }
        exm = matrix_mulit(exm, exm);
        gg >>= 1;
    }
     return ans.m[0][0];
} 
int n; 
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==-1)
         break;    
        printf("%d
",matrix_quick(n));
    }
    return 0;
}