poj3070 Fibonacci（矩阵乘法初学）

最新推荐文章于 2021-09-24 14:28:56 发布

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#矩阵乘法

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本文介绍如何使用矩阵快速幂算法高效地计算Fibonacci数列中任意项的最后四位数字，避免了传统的递归或迭代方法可能导致的性能瓶颈。

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Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9972		Accepted: 7114

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

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代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define mod 10000

struct mat{
   int A[3][3];
}s;

mat matmul(mat x,mat y)       //x*y  
{
	mat t;
    t.A[0][0]=(x.A[0][0]*y.A[0][0]+x.A[0][1]*y.A[1][0])%mod;
    t.A[0][1]=(x.A[0][0]*y.A[0][1]+x.A[0][1]*y.A[1][1])%mod;

    t.A[1][0]=(x.A[1][0]*y.A[0][0]+x.A[1][1]*y.A[1][0])%mod;
    t.A[1][1]=(x.A[1][0]*y.A[0][1]+x.A[1][1]*y.A[1][1])%mod;
    return t;
}

mat matpow(mat ss,int k)      //ss^k
{
	if(k==1) return s;
	ss=matpow(ss,k/2);

	if(k%2==0)
	{
		return matmul(ss,ss);
	}
	else
		return matmul(matmul(ss,ss),s);
}

int main()
{
	int i,j,n;

	while(~scanf("%d",&n)&&n!=-1)
	{
        if(n==0)
		{
			printf("0\n");
			continue;
		}

		s.A[0][0]=s.A[0][1]=s.A[1][0]=1;
		s.A[1][1]=0;

        s=matpow(s,n);
		printf("%d\n",s.A[0][1]);
	}
	return 0;
}