HDU2604Queuing

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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

Sample Output

  
  
   
   6
2
1
  
  
  
  
   
   分析： 题目要求不含101 和111 的串
  
  
  
  
   
   设f[n]为长度为n串符合条件的个数
  
  
  
  
   
   则很明显在长度为（n-1）并且符合条件的串后面加上一个0一定符合
  
  
  
  
   
   如果在长度为n-1的串后面加上一个1的话我们得考虑n-1的串结尾的元素
  
  
  
  
   
   如果是00的话我们可以看做是长度为n-3的串加上100 
  
  
  
  
   
   如果是10的话我们可以看做长度为n-4的串加上1100
  
  
  
  
   
   因此f[n]=f[n-1]+f[n-3]+f[n-4];
  
  
  
  
   
   方法一：直接递推（耗时多可能超时）
  
  
  
  
   
   
   
   #include <iostream>
#include <cstdio>
using namespace std;
int n,M;
int main()
{
    int n,m;
    int a[4];
    while(~scanf("%d%d",&n,&M)){
        a[0]=1;
        a[1]=2;
        a[2]=4;
        a[3]=6;
        for(int i=4;i<=n;i++){
            a[4]=(a[0]+a[1]+a[3])%M;
             a[0]=a[1];
             a[1]=a[2];
             a[2]=a[3];
             a[3]=a[4];
        }
        if(n>=4)
          cout<<a[4]%M<<endl;
        else
            cout<<a[n]%M<<endl;
    }
    return 0;
}
   
   
方法二：（矩阵加速）
  
  
  
  
   
   
   
   #include <iostream>
#include <cstdio>
using namespace std;
int n,M;
struct matrax
{
    int m[4][4];
};
matrax A={
  1,0,1,1,
  1,0,0,0,
  0,1,0,0,
  0,0,1,0
};
matrax E;
void init()
{
   for(int i=0;i<4;i++)
    for(int j=0;j<4;j++)
     E.m[i][j]=(i==j);
}
matrax multi(matrax a,matrax b)
{
    matrax c;
    for(int i=0;i<4;i++){
        for(int j=0;j<4;j++){
            c.m[i][j]=0;
            for(int k=0;k<4;k++)
                c.m[i][j]+=a.m[i][k]*b.m[k][j]%M;
        c.m[i][j]%=M;
        }
    }
    return c;
}
matrax power(matrax A,int k)
{
    matrax ans=E,p=A;
    while(k){
        if(k&1){
            ans=multi(ans,p);
            k--;
        }
        k>>=1;
        p=multi(p,p);
    }
    return ans;
}
int main()
{
    init();
    int a[4]={1,2,4,6};
    while(cin>>n>>M){
        matrax ans=power(A,n-3);
        int x=0;
        for(int i=0;i<4;i++)
            x+=(ans.m[0][i]*a[4-i-1])%M;
        /*for(int i=0;i<4;i++){
            for(int j=0;j<4;j++)
                cout<<ans.m[i][j]<<" ";
            cout<<endl;
        }*/
        cout<<x%M<<endl;
    }
    return 0;
}