Poj 3070 Fibonacci_牛客博客

Time Limit: 1000MS        Memory Limit: 65536K
Total Submissions: 26063        Accepted: 17394

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Poj 3070 Fibonacci

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 +
Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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:

题意：

其实就是求第x位斐波那契数列（记得mod10000），不过题目给你介绍了矩阵的相关知识，以及矩阵与斐波那契数列的联系，就是让你用矩阵快速幂来求斐波那契数列。

题解：

斐波那契数列的博客
我在这个博客里详细介绍了三种斐波那契数列的求法，其中就有矩阵快速幂的方法
所以我在这就不详细介绍，简单说说：
mul（a，b）是用来进行矩阵a与b的乘法运算
pow(mat a,ll n)//快速幂求矩阵a的n次幂

代码：

#include <iostream>  
#include <cstddef>  
#include <cstring>  
#include <vector>  
using namespace std;  
typedef long long ll;  
const int mod=10000;  
typedef vector<ll> vec;  
typedef vector<vec> mat;  
mat mul(mat &a,mat &b) 
{  
    mat c(a.size(),vec(b[0].size()));  
    for(int i=0; i<2; i++)  
    {  
        for(int j=0; j<2; j++)  
        {  
            for(int k=0; k<2; k++)  
            {  
                c[i][j]+=a[i][k]*b[k][j];//矩阵的运算法则 
                c[i][j]%=mod;  
            }  
        }  
    }  
    return c;  
}  
mat pow(mat a,ll n)  
{  
    mat res(a.size(),vec(a.size()));  
    for(int i=0; i<a.size(); i++)  
        res[i][i]=1;  
    while(n)  
    {  
        if(n&1) res=mul(res,a);  
        a=mul(a,a);  
        n/=2;  
    }  
    return res;  
}  
ll solve(ll n)  
{  
    mat a(2,vec(2));  
    a[0][0]=1;  
    a[0][1]=1;  
    a[1][0]=1;  
    a[1][1]=0;  
    a=pow(a,n);  
    return a[0][1];
}  
int main()  
{  
    ll n;  
    while(cin>>n&&n!=-1)  
    {  
        cout<<solve(n)<<endl;  
    }  
    return 0;  
}