nyoj_148_fibonacci数列_矩阵快速幂

Posted 2020-08-12 多一份不为什么的坚持

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[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;
}