poj3070 Fibonacci（矩阵快速幂）

最新推荐文章于 2021-09-24 14:28:56 发布

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本文介绍了一种使用矩阵快速幂算法高效计算斐波那契数列的方法，并提供了一个C++实现示例。该算法能显著减少计算过程中的重复运算，适用于大整数的斐波那契数计算。

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思路：矩阵快速幂的入门题

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct Mat
{
	int a[2][2];   //矩阵大小
};
int n;
const int mod = 10000;
Mat mul(Mat a,Mat b)
{
	Mat t;
	memset(t.a,0,sizeof(t.a));
	for(int i = 0;i<n;i++)
	{
		for(int k = 0;k<n;k++)
		{
			if(a.a[i][k])
              for(int j = 0;j<n;j++)
			  {
				  t.a[i][j]+=a.a[i][k]*b.a[k][j];
				  if(t.a[i][j]>=mod)
					  t.a[i][j]%=mod;
			  }
		}
	}
	return t;
}
Mat expo(Mat p,int k)
{
	if(k==1)return p;
	Mat e;
	memset(e.a,0,sizeof(e.a));
	for(int i = 0;i<n;i++)   //初始化单位矩阵  
		e.a[i][i]=1;
    if(k==0)return e;
	while(k)
	{
		if(k&1)
			e = mul(p,e);
		p = mul(p,p);
		k>>=1;
	}
	return e;
}
int main()
{
    int k;
	n=2;
	while(scanf("%d",&k)!=EOF && k!=-1)
	{
        if(k==0)
			printf("0\n");
		else
		{
			Mat p;
			p.a[0][0]=p.a[0][1]=p.a[1][0]=1;
			p.a[1][1]=0;
			Mat ans = expo(p,k);
			/*for(int i = 0;i<n;i++,printf("\n"))
				for(int j = 0;j<n;j++)
					printf("%d ",ans.a[i][j]);*/
			printf("%d\n",ans.a[0][1]);
		}
	}
}

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., printF_n mod 10000).

Sample Input