hdu2604之矩阵快速幂

Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2003    Accepted Submission(s): 938

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. 

  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

   
   
   
   

    
    
    
    3 8
4 7
4 8
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    6
2
1
   
   
   
   

 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#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){
		/*for(int i=3;i<=n;++i){//推出递推公式了就可以用矩阵快速幂了 
			dp[i][0]=dp[i-1][3];
			dp[i][1]=(dp[i-1][0]+dp[i-1][3])%m;
			dp[i][2]=(dp[i-1][1]+dp[i-1][2])%m;
			dp[i][3]=dp[i-1][2];
		}*/
		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;
			//cout<<(dp[n][0]+dp[n][1]+dp[n][2]+dp[n][3])%m<<endl;
		}
	}
	return 0;
}