POJ 3070 Fibonacci 快速矩阵幂

最新推荐文章于 2019-03-28 17:11:09 发布

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本文介绍了一种通过快速矩阵幂的方法来高效计算斐波那契数列第n项的最后四位数字。这种方法避免了直接递归所带来的大量重复计算，大大提高了算法效率。

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Language:

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13723		Accepted: 9720

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

Sample Output

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

题目链接：http://poj.org/problem?id=3070

<span style="font-size:24px;">#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int mod=10000;
int n;
struct My{
int cun[2][2];
My operator*(const My m)const{
My temp;
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
 temp.cun[i][j]=0;
    for(int k=0;k<2;k++)
    {
        temp.cun[i][j]+=cun[i][k]*m.cun[k][j]%mod;
        temp.cun[i][j]%=mod;
    }
}
}
return temp;
}
};

int Pow(My &m,int k){
    My ans;
    memset(ans.cun,0,sizeof(ans.cun));
    for(int i = 0;i <2;i++)
        ans.cun[i][i] = 1;
    while(k){
        if(k&1){ ans = ans*m;}
        k >>= 1;
        m = m*m;
    }
    return ans.cun[0][0]%mod;
}

int  main(){
while(cin>>n&&n!=-1)
{
    if(n==0){cout<<0<<endl;continue;}
    My m;
    m.cun[0][0]=1,m.cun[1][0]=1,m.cun[0][1]=1,m.cun[1][1]=0;
int res=Pow(m,n-1);
cout<<res<<endl;
}
return 0;}</span><span style="font-size: 12pt;">
</span>

这是一个快速矩阵幂的模板题，主要有一个实现矩阵的结构体，需要重载一下结构体的乘法（根据矩阵乘法的原理），然后利用类似二分的原理，如：计算n的八次方分成了

计算n的平方，n的平方乘上n的平方。。。依次列推。可得答案。