HDU2604- Queuing(矩阵高速幂优化)

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 

 

记录dp[i][4]为队伍长度为i时的四种状态mm mf fm ff

那么(mm)dp[i][0] = dp[i-1][0] + dp[i-1][2];

(mf)dp[i][1] = dp[i-1][0] ;

(fm)dp[i][2] = dp[i-1][1] + dp[i-1][3] ;

(ff)dp[i][3] = dp[i-1][1] ;

那么就可以使用矩阵优化

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
struct node{
    int a[5][5] ;
    int n ;
};
node mul(node p,node q,int m)
{
    int i , j , k ;
    node s ;
    s.n = p.n ;
    for(i = 0 ; i < p.n ; i++)
        for(j = 0 ; j < p.n ; j++)
        {
            s.a[i][j] = 0 ;
            for(k = 0 ; k < p.n ; k++)
                s.a[i][j] = ( s.a[i][j] + p.a[i][k] * q.a[k][j] ) % m ;
        }
    return s ;
}
node pow(node p,int k,int m)
{
    if( k == 1 )
        return p ;
    node s = pow(p,k/2,m) ;
    s = mul(s,s,m) ;
    if( k%2 )
        s = mul(s,p,m) ;
    return s ;
}
int main()
{
    int i , j , l , m ;
    node p , s ;
    p.n = 4 ;
    memset(p.a,0,sizeof(p.a)) ;
    p.a[0][0] = p.a[0][1] = 1 ;
    p.a[1][2] = p.a[1][3] = 1 ;
    p.a[2][0] = 1 ;
    p.a[3][2] = 1 ;
    while( scanf("%d %d", &l, &m) != EOF )
    {
        if(l == 0 )
        {
            printf("1\n") ;
            continue ;
        }
        else if( l == 1 )
        {
            printf("2\n") ;
            continue ;
        }
        else if( l == 2 )
        {
            printf("4\n") ;
            continue ;
        }
        s = pow(p,l-2,m) ;
        int ans = 0 ;
        for(i = 0 ; i < 4 ; i++)
            for(j = 0 ; j < 4 ; j++)
                ans = ( ans + s.a[i][j] ) % m ;
        printf("%d\n", ans) ;
    }
    return 0;
}