poj3070 Fibonacci(矩阵快速幂)

最新推荐文章于 2021-07-08 09:30:58 发布

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最新推荐文章于 2021-07-08 09:30:58 发布

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Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11520		Accepted: 8188

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:

#include <iostream>
#include <cstdio>

using namespace std;

const int mod = 10000;
struct matrix {  //用一个结构体来存矩阵
	int node[2][2];
}ans, tem;

matrix muti(matrix a, matrix b) {  //矩阵a乘b返回结果矩阵
	matrix temp;
	for (int i = 0; i < 2; i++) {
		for (int j = 0; j < 2; j++) {
			temp.node[i][j] = 0;
			for (int k = 0; k < 2; k++) {
				temp.node[i][j] = (temp.node[i][j] + (a.node[i][k] * b.node[k][j])) % mod;
			}
		}
	}
	return temp;
}

int quick_m(int b) {
	ans.node[0][0] = ans.node[1][1] = 1;  //ans初始化为单位矩阵
	ans.node[0][1] = ans.node[1][0] = 0;
	tem.node[0][0] = tem.node[0][1] = tem.node[1][0] = 1;
	tem.node[1][1] = 0;
	while (b) {   //快速幂使用到矩阵乘法中
		if (b & 1) {
			ans = muti(ans, tem);
		}
		tem = muti(tem, tem);
		b >>= 1;
	}
	return ans.node[0][1];  //注意返回的是F(n)
}

int main()
{
	int n;
	while (~scanf("%d", &n) && n != -1) {
		printf("%d\n", quick_m(n));
	}
	return 0;
}