【poj 3070】Fibonacci

最新推荐文章于 2021-09-24 14:28:56 发布

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本文介绍了一种高效计算斐波那契数列第n项最后四位数字的方法，利用矩阵快速幂来减少计算复杂度，并提供了一个C++实现示例。

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D - Fibonacci

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
#define Mod 10000
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
struct Mat
{
	int a[4][4];
	int h,w;
}pr,ans;
void init()
{
	ans.a[2][1]=ans.a[1][2]=0;  //初始化过渡矩阵 
	ans.h=ans.w=2;
	for(int i=1;i<=2;i++)
	ans.a[i][i]=1;
	pr.a[1][1]=pr.a[1][2]=pr.a[2][1]=1; //初始化单位矩阵 
	pr.a[2][2]=0;
	pr.h=pr.w=2;
}
Mat Mat_Mul(Mat x,Mat y)  //矩阵相乘 
{
	Mat t;
	CLR(t.a,0);
	t.h=x.h;
	t.w=y.w;
	for(int i=1;i<=x.h;i++)
	{
		for(int j=1;j<=x.w;j++)
		{
			if(x.a[i][j]==0) continue;
			for(int k=1;k<=y.w;k++)
			t.a[i][k]=(t.a[i][k]+x.a[i][j]*y.a[j][k]%Mod)%Mod;
		}
	}
	return t;
}
void Mat_mod(int n)  //矩阵快速幂 
{
	while(n)
	{
		if(n&1)
		ans=Mat_Mul(pr,ans);
		pr=Mat_Mul(pr,pr);
		n>>=1;
	}
}
int main()
{
    int n;
    while(~scanf("%d",&n)&&n!=-1)
    {
    	init();
    	Mat_mod(n);
    	printf("%d\n",ans.a[1][2]%Mod);
	}
	return 0;
}