HDU 2604 Queuing

Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4638    Accepted Submission(s): 2045

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

 

Author

WhereIsHeroFrom

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

Recommend

lcy

 

对于长度为n的队列q(n)，做以下讨论。

图中 ，".....m"表示以字符'm'结尾的queue，其它类似。E_q表示E_queue。

 

定义f(n)为队列q(n)之中满足E_queue的队列个数。

可得f(n)=f(n-1)+f(n-3)+f(n-4)

转换为矩阵，运用矩阵快速幂运算进行求解。

即，

n>=4,

 

 

 

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <functional>
#include <string>
#include <cstring>
#include <queue>
#include <set>
#include <cmath>
#include <cstdio>
using namespace std;
#define IOS ios_base::sync_with_stdio(false)]
typedef long long LL;
const int INF=0x3f3f3f3f;

const int maxn=6;
int modnum;
typedef struct matrix{
    int v[maxn][maxn];
    void init(){memset(v,0,sizeof(v));}
}M;
M mul_mod(const M &a,const M &b,int L,int m,int n)
{
    M c; c.init();
    for(int i=0;i<L;i++){
        for(int j=0;j<n;j++){
            for(int k=0;k<m;k++)//注意j，k范围
                c.v[i][j]+=a.v[i][k]*b.v[k][j]%modnum;
            c.v[i][j]%=modnum;
        }
    }
    return c;
}
M power(M x,int L,int p)
{
    M tmp; tmp.init();
    for(int i=0;i<L;i++)
        tmp.v[i][i]=1;
    while(p){
        if(p&1) tmp=mul_mod(x,tmp,L,L,L);
        x=mul_mod(x,x,L,L,L);
        p>>=1;
    }
    return tmp;
}
int main()
{
    int L;
    M a,b;
    while(scanf("%d%d",&L,&modnum)!=EOF){
        b.init();
        b.v[0][0]=9%modnum;
        b.v[1][0]=6;
        b.v[2][0]=4;
        b.v[3][0]=2;
        if(L==0){printf("1
"); continue;}
        if(L<5){printf("%d
",b.v[4-L][0]); continue;}
        a.init();
        a.v[0][0]=a.v[0][2]=a.v[0][3]=a.v[1][0]=a.v[2][1]=a.v[3][2]=1;
        a=power(a,4,L-4);
        a=mul_mod(a,b,4,4,1);
        printf("%d
",a.v[0][0]);
    }
}

网上有介绍用trie图得到递推式的方法，根节点为字符串的最后一个字符，根据前一个字符进行状态转移。

不是非常理解，暂且记录下来。