HDU

　　　　Queuing

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6639 Accepted Submission(s): 2913

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

8
7
8

Sample Output

6
2
1

题意

给定一个队伍长度k，和mod，求队伍有多少种排列可能。其中，队伍排列要求：不能出现 fff 或者 fmf

解题思路

类似递推思路， 1.若最后一个为m，则无论前一个为什么情况都可以，sum+=dp[i-1]

若最后一个为f，则　　{

　　　　　　　　　　　　　2.　若前一位为m，则再之前一位必定为m，此时队列为mmf,此时可同第一种情况，由mmf前一位决定，此时sum+=dp[i-3]

　　　　　　　　　　　　　3.　若前一位为f,则队伍要符合题意之前依旧只能是mm,原理同第二种情况，此时队列为mmff，由mmff前一位决定，此时sum+=dp[i-4]

　　　　　　　　　　}

得递推式，f(n)=f(n-1)+f(n-3)+f(n-4);

之后直接矩阵快速幂即可

则 f(n) 　　 1 0 1 1 　 f(n-1)

　 f(n-1) 　 1 0 0 0　　 f(n-2)

　 f(n-2) 0 1 0 0　 f(n-3)

f(n-3) 0 0 1 0 　　f(n-4)

手推枚举前4项，得f(1)=2 f(2)=4 f(3)=6 f(4)=9

具体实现看代码吧

#include<bits/stdc++.h>
using namespace std;
const int maxn=50;
typedef long long ll;
#define mod(x) ((x)%MOD)
ll MOD;
//2 4 6 9
struct mat
{
    int m[maxn][maxn];
    mat(){
        memset(m,0,sizeof(m));
    }
}unit;
mat operator*(mat a,mat b)
{
    mat ret;
    ll x,n=4;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            x=0;
            for(int k=0;k<n;k++)
            {
                x+=mod((ll)a.m[i][k]*b.m[k][j]);
            }
            ret.m[i][j]=mod(x);
        }
    }
    return ret;
}
void iint()
{
    for(int i=0;i<maxn;i++)
    {
        unit.m[i][i]=1;
    }
    return ;
}
mat pow1(mat a,ll n)
{
    mat ret=unit;
    while(n)
    {
        if(n&1)
        {
            n--;ret=ret*a;
        }
        n>>=1;
        a=a*a;
    }
    return ret;
}
int main()
{
    ll k;
    mat b;
    b.m[0][0]=9; b.m[1][0]=6; b.m[2][0]=4;  b.m[3][0]=2;   
    //f(1)对于f(n)来说是f(n-4)，这四项写反，查错了好久，哭
    while(cin>>k>>MOD)
    {
        if(k<=4) {  cout<<b.m[4-k][0]%MOD<<endl;continue;}
        iint(); //构建单位阵
        mat a;
        a.m[0][0]=a.m[0][2]=a.m[0][3]=a.m[1][0]=a.m[2][1]=a.m[3][2]=1;
        //a.m[0][1]=a.m[1][1]=a.m[1][2]=a.m[1][3]=a.m[2][0]=a.m[2][2]=a.m[2][3]=a.m[3][0]=a.m[3][1]=a.m[3][3]=0;
        a=pow1(a,k-4); //进行k-4次快速幂即可
        a=a*b;
        /*for(int i=0;i<4;i++)
        { //这是查看矩阵的= =
            for(int j=0;j<4;j++)
            {
                if(j+1==4) cout<<a.m[i][j]<<endl;
                else cout<<a.m[i][j]<<" ";
            }
        }*/
        cout<<mod(a.m[0][0])<<endl;
    }
    return 0;
}