Poj 3070 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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本文介绍了一种高效计算斐波那契数列中指定项最后四位数字的方法，使用矩阵快速幂算法来减少计算复杂度，并提供了完整的C++代码实现。

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Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16617		Accepted: 11637

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

0
9
999999999

1000000000

-1

Sample Output

AC代码：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const int MOD = 10000;
struct mut{
	int at[4][4];
};
int q;
mut d;

//两个矩阵相乘 
mut mal(mut a, mut b){
	mut c;
	memset(c.at, 0, sizeof(c.at));
	for(int i = 1; i <= q; i++){
		for(int j = 1; j <= q; j++){
			for(int k = 1; k <= q; k++){
				c.at[i][j] += a.at[i][k] * b.at[k][j];
				if(c.at[i][j] >= MOD) c.at[i][j] %= MOD;
			}
		}
	}
	return c;
} 

//求n次幂矩阵 
mut expo(mut a, int k){
	mut e;
	memset(e.at, 0, sizeof(e.at));
	for(int i = 1; i <= q; i++) e.at[i][i] = 1;
	while(k){
		if(k & 1) e = mal(e, a);
		a = mal(a, a);
		k >>= 1;
	}
	return e;
}

int main(){
	q = 2;
	int n;
	d.at[1][1] = 1;
	d.at[1][2] = 1;
	d.at[2][1] = 1;
	d.at[2][2] = 0;
//	mut ans = expo(d, 3);
//	for(int i = 1; i <= 2; i++){
//		for(int j = 1; j<= 2; j++){
//			printf("%d ", ans.at[i][j]);
//		}
//		printf("\n");
//	}

	while(scanf("%d", &n) != EOF && n != -1){
		if(n == 0) printf("0\n");
		else if(n == 1)	printf("1\n");
		else{
			mut ans = expo(d, n);
			printf("%d\n", ans.at[1][2]);
		} 
	}
	return 0;
}