描述

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.

 

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:

.

 

输入

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.

输出

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

样例输入

091000000000-1

样例输出

0346875

来源

POJ

上传者

hzyqazasdf

解题思路：

一下午就学了这一个算法，有很多细节总是花很长时间才理解。感觉自己学习算法效率好低啊。

之前刚接触斐波那契数列，想找一个更高效的方法来求，当时看到了，却根本不懂。原来这就是矩阵快速幂。。。从此有了求斐波那契数列更好的方法

既然整数求幂可以用快速幂来求，那么矩阵的幂同样也可以啊。

#include <iostream>#include <cstdio>#include <cstring>#define mod 10000using namespace std;struct matrix{    int m[2][2];};matrix base,ans;void init(int n){//只初始化base和ans(单位矩阵)    memset(base.m,0,sizeof(base.m));    memset(ans.m,0,sizeof(ans.m));    for(int i=0;i<2;i++){        ans.m[i][i]=1;    }    base.m[0][0]=base.m[0][1]=base.m[1][0]=1;}matrix multi(matrix a,matrix b){    matrix t;    for(int i=0;i<2;i++){        for(int j=0;j<2;j++){            t.m[i][j]=0;            for(int k=0;k<2;k++){                t.m[i][j]=(t.m[i][j]+a.m[i][k]*b.m[k][j])%mod;            }        }    }    return t;}int fast_matrix(int n){    while(n){        if(n&1){            ans=multi(ans,base);        }        base=multi(base,base);        n>>=1;    }    return ans.m[1][0];}int main(){    int n;    while(~scanf("%d",&n) && n!=-1){        init(n);        printf("%d\n",fast_matrix(n));    }    return 0;}

nyoj_148_fibonacci数列(二)_矩阵快速幂