HDU - 2604 Queuing （矩阵快速幂）

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我是靠谱客的博主爱笑樱桃，最近开发中收集的这篇文章主要介绍HDU - 2604 Queuing （矩阵快速幂），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Description

Queues and Priority Queues are data structures which are known to most computer scientists. The Queue occurs often in our daily life. There are many people lined up at the lunch time.

Now we define that ‘f’ is short for female and ‘m’ is short for male. If the queue’s length is L, then there are 2 ^L numbers of queues. For example, if L = 2, then they are ff, mm, fm, mf . If there exists a subqueue as fmf or fff, we call it O-queue else it is a E-queue.
Your task is to calculate the number of E-queues mod M with length L by writing a program.

Input

Input a length L (0 <= L <= 10 ⁶) and M.

Output

Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.

Sample Input

Sample Output

题意：n个人排队，f表示女，m表示男，形成的2^n个序列中，有多少个序列不包含子串‘fmf’和‘fff’.

思路：用f(n)表示n个人满足条件的结果，那么如果最后一个人是m的话，那么前n-1个满足条件即可，就是f(n-1)；

如果最后一个是f那么这个还无法推出结果，那么往前再考虑一位：那么后三位可能是：mmf, fmf, mff, fff，其中fff和fmf不满足题意所以我们不考虑，但是如果是

mmf的话那么前n-3可以找满足条件的即：f(n-3)；如果是mff的话，再往前考虑一位的话只有mmff满足条件即：f(n-4)

所以f(n)=f(n-1)+f(n-3)+f(n-4)，为了达到效率的要求我们应用矩阵快速幂

构造一个矩阵：借用一张图来表达

写出一次相乘的矩阵结果是： f(n-1)+f(n-3)+f(n-4)

f(n-1)

f(n-2)

f(n-3)，那么就可以递推出结果是在矩阵的第一项

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 4;

struct Matrix {
	int v[maxn][maxn];
	Matrix() {
		memset(v, 0, sizeof(v));
	}
};
int n, m;

Matrix Mul(Matrix a, Matrix b) {
	Matrix c;
	for (int i = 0; i < maxn; i++) 
		for (int j = 0; j < maxn; j++) 
			for (int k = 0; k < maxn; k++)
				c.v[i][j] = (c.v[i][j]+a.v[i][k]*b.v[k][j])%m;
	return c;
}

Matrix powMul(Matrix a, int k) {
	Matrix tmp;
	for (int i = 0; i < maxn; i++)
		tmp.v[i][i] = 1;
	while (k) {
		if (k & 1)
			tmp = Mul(a, tmp);
		k >>= 1;
		a = Mul(a, a);
	}
	return tmp;
}

int main() {
	Matrix a, b, c;
	a.v[0][0] = 9;
	a.v[1][0] = 6;
	a.v[2][0] = 4;
	a.v[3][0] = 2;
	b.v[0][0] = b.v[0][2] = b.v[0][3] = b.v[1][0] = b.v[2][1] = b.v[3][2] = 1;
	while (scanf("%d%d", &n, &m) != EOF) {
		if (n <= 4) {
			if (n == 0)
				printf("0n");
			else printf("%dn", a.v[4-n][0]%m);
		}
		else {
			c = powMul(b, n-4);
			c = Mul(c, a);
			printf("%dn", c.v[0][0]%m);
		}
	}
	return 0;
}