hpuvj【3070】Fibonacci（矩阵快速幂求模）

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本文介绍了一种利用矩阵快速幂算法高效计算斐波那契数列第n项最后四位数字的方法。通过定义特定的2x2矩阵并利用矩阵乘法及快速幂运算，文章提供了一个C++实现示例。

Fibonacci

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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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

一道涉及到矩阵快速幂的知识的题本来快速幂没学好今天还专门说这个了，所以看了半天才略懂了一些，代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct Mat//定义矩阵
{
	int a[2][2];
	void init()
	{
		memset(a,0,sizeof(a));
		a[0][0]=a[1][1]=1;
	 } 
};
Mat mul(Mat a,Mat b)//矩阵乘法 
{
	Mat ans;
	ans.init();
	for(int i=0;i<2;i++)
	{
		for(int j=0;j<2;j++)
		{
			ans.a[i][j]=0;
			for(int k=0;k<2;k++)
            {
                ans.a[i][j]+=a.a[i][k]*b.a[k][j];
            }
            ans.a[i][j]%=10000;
		}
	}
	return ans;
}
Mat power(Mat a,int num)//矩阵快速幂模板 
{
	Mat ans;
	ans.init();
	while(num)
	{
		if(num%2==1)
		{
			ans=mul(ans,a);
		}
		num/=2;
		a=mul(a,a);
	}
	return ans;
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF&&n!=-1)
	{
		if(n==0)
		{
			printf("0\n");
			continue;
		}
		Mat a,ans;
		a.a[0][1]=a.a[0][0]=a.a[1][0]=1;
		a.a[1][1]=0;
		ans=power(a,n);
		printf("%d\n",ans.a[0][1]%10000);
	}
	return 0;
}