课程练习三-1009-problem I

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我是靠谱客的博主含糊舞蹈，最近开发中收集的这篇文章主要介绍课程练习三-1009-problem I，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Problem 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. <center><img src=../../../data/images/C154-1005-1.jpg> </center>   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 2L 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

  
  
   
   6
2

   
   1
   
   

   
   题意：这个怎么说，就是找 递推公式，计算呗！
   
   

   
   思路：
   
   矩阵快速幂（可以先拿Fibonacci数列试手，网上有Fibonacci数列快速幂的解析，自己查）
   
   主要是找递推公式，
   
   f(n)=f(n-1)+g;
   
   f(n)=f(n-1)+f(n-3)+g';
   
   f(n)=f(n-1)+f(n-3)+f(n-4);

   
   SO f(n)=f(n-1)+f(n-3)+f(n-4)位最后的公式。
   
   剩下的基本套模板就行了。
   
   

   
   AC代码：
   
   #include<iostream>  
#include<algorithm>  
#include<stdlib.h>
#define INF 99999999  
using namespace std;


const int MAX = 4;
int array[MAX][MAX], sum[MAX][MAX];


void MatrixMult(int a[MAX][MAX], int b[MAX][MAX], int &mod){
	int c[MAX][MAX] = { 0 };
	for (int i = 0; i<MAX; ++i){
		for (int j = 0; j<MAX; ++j){
			for (int k = 0; k<MAX; ++k){
				c[i][j] += a[i][k] * b[k][j];
			}
		}
	}
	for (int i = 0; i<MAX; ++i){
		for (int j = 0; j<MAX; ++j)a[i][j] = c[i][j] % mod;
	}
}


int MatrixPow(int k, int &mod){
	for (int i = 0; i<MAX; ++i){
		for (int j = 0; j<MAX; ++j)sum[i][j] = (i == j);
	}
	array[0][0] = array[0][1] = array[0][2] = 0, array[0][3] = 1;
	array[1][1] = array[1][2] = 0, array[1][0] = array[1][3] = 1;
	array[2][0] = array[2][3] = 0, array[2][1] = array[2][2] = 1;
	array[3][0] = array[3][1] = array[3][3] = 0, array[3][2] = 1;
	while (k){
		if (k & 1)MatrixMult(sum, array, mod);
		MatrixMult(array, array, mod);
		k >>= 1;
	}
	int ans = 0;
	for (int i = 0; i<MAX; ++i)ans = (ans + sum[i][0] + sum[i][1] + sum[i][2] + sum[i][3]) % mod;
	return ans;
}


int main(){
	int n, m;
	while (cin >> n >> m){
		if (n == 0)cout << 0 << endl;
		else if (n == 1)cout << 2 % m << endl;
		else if (n == 2)cout << 4 % m << endl;
		else{
			cout << MatrixPow(n - 2, m) << endl;
		}
	}
	system("pause");
	return 0;
}

   
   

   
   感想：很长时间没复习快速幂了，这次复习一下。