nyoj_148_fibonacci数列(二)_矩阵快速幂

fibonacci数列(二)

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

 

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

0
9
1000000000
-1

样例输出

0
34
6875

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