fibonacci数列(二)_矩阵快速幂

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

【题意】

斐波那契数列可以用矩阵来求

当求第非常大的一个斐波那契数的后几位时我们可以用矩阵快速幂求解了。

#include<iostream>
#include<stdio.h>
#include<vector>
#include<string.h>
using namespace std;
typedef vector<int>vec;
typedef vector<vec>mat;
const int N=10000;
mat mul(mat a,mat b)//求两个矩阵的乘积
{
    mat c(a.size(),vec(b[0].size()));
    for(int i=0;i<a.size();i++)
    {
        for(int k=0;k<b.size();k++)
        {
            for(int j=0;j<b[0].size();j++)
            {
                c[i][j]=(c[i][j]+a[i][k]*b[k][j])%N;
            }
        }
    }
    return c;
}
mat get_ans(mat a,int n)//矩阵的快速幂
{
    mat b(a.size(),vec(a.size()));
    for(int i=0;i<a.size();i++)
    {
        b[i][i]=1;
    }
    while(n>0)
    {
        if(n&1) b=mul(b,a);
        a=mul(a,a);
        n>>=1;
    }
    return b;
}
int main()
{
    long long int n;
    while(~scanf("%lld",&n),n>=0)
    {
        if(n==-1) break;
        mat a(2,vec(2));
        a[0][0]=1,a[0][1]=1;
        a[1][0]=1,a[1][1]=0;
        a=get_ans(a,n);
        printf("%d
",a[1][0]);
    }
    return 0;
}