【POJ】3070 - Fibonacci（矩阵快速幂）

FishWang

发布于 2025-08-26 19:57:49

11100

代码可运行

运行总次数：0

代码可运行

点击打开题目

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12951		Accepted: 9210

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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

这道题大概就是矩阵快速幂的基本模板了。注释说的清楚。

注意下矩阵相乘的下标就行了。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define MOD 10000
struct Matrix
{
	__int64 a[4][4];		//矩阵
	int h,w;		//行、列 
}pr,ans;
void init()		//初始化矩阵 
{
	ans.a[2][1] = ans.a[1][2] = 0;
	ans.h = ans.w = 2;
	for (int i = 1 ; i <= 2 ; i++)		//初始化单位矩阵 
		ans.a[i][i] = 1;
	pr.a[1][1] = pr.a[1][2] = pr.a[2][1] = 1;		//初始化初始矩阵 
	pr.a[2][2] = 0;
	pr.w = 2;
	pr.h = 2;
}
Matrix Matrix_multiply(Matrix x , Matrix y)		//矩阵乘法 
{
	Matrix t;
	CLR(t.a,0);
	t.h = x.h;		//新矩阵的行数为x的行数 
	t.w = y.w;		//新矩阵的列数为y的列数 
	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 Matrix_mod(int n)		//矩阵快速幂
{
	while (n)
	{
		if (n & 1)
			ans = Matrix_multiply(pr , ans);
		pr = Matrix_multiply(pr , pr);
		n >>= 1;
	}
}
int main()
{
	int n;
	while (~scanf ("%d",&n) && n != -1)
	{
		init();
		Matrix_mod(n);
		printf ("%I64d\n",ans.a[1][2] % MOD);
	}
	return 0;
}

本文参与腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。

原始发表：2025-08-26，如有侵权请联系 cloudcommunity@tencent.com 删除

digits