Fibonacci POJ - 3070 （矩阵快速幂的模板）-优快云博客

本文链接：https://blog.youkuaiyun.com/obsorb_knowledge/article/details/80329777

本文介绍了一种利用矩阵快速幂的方法来高效计算斐波那契数列中任意一项的最后四位数字，适用于大整数计算场景，并提供了一个具体的C++实现案例。

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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_nare 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:

题意：就是求 Fibonacci 数列的第n项的数；

思路：矩阵快速幂，最重要的就是构造式子，

如：在斐波那契数列之中

fi[i] = 1*fi[i-1]+1*fi[i-2]

fi[i-1] = 1*f[i-1] + 0*f[i-2];

即

所以

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long
#define mod 10000
struct node
{
	ll a[2][2];
};
ll n;

node s_cheng(node tt,node kk)
{
	int i,j,k;
	node c;
	c.a[0][0] = c.a[0][1] = c.a[1][0] = c.a[1][1] = 0;
	for(i=0;i<2;i++)
		for(j = 0;j<2;j++)
			for(k = 0;k<2;k++)
				c.a[i][j] = (c.a[i][j] + tt.a[i][k] * kk.a[k][j])%mod; 
	return c;
}

node jz_pow(node tt,ll n)
{
	node res;
	res.a[0][0] = res.a[1][1] = 1;
	res.a[0][1] = res.a[1][0] = 0;
	while(n)
	{
		if(n&1) res = s_cheng(res,tt);
		n>>=1;
		tt = s_cheng(tt,tt);
	}
	return res;
}
int main()
{
	while(~scanf("%lld",&n)&&n!=-1)
	{
		if(n==0)
		{
			printf("0\n");
			continue;
		} 
		node tt,res;
		tt.a[0][0] = tt.a[0][1] = tt.a[1][0] = 1;
		tt.a[1][1] = 0;
		res.a[0][0] = 1;
		res.a[1][0] = 0;
		tt = jz_pow(tt,n-1);
		res = s_cheng(tt,res);
		printf("%lld\n",res.a[0][0]); 
	}
	return 0;
}