HDU 2604-Queuing

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 84 74 8

 

Sample Output

621

F[0] = 0 , F[1] = 2 , F[2] = 4 , F[3] = 6 , F[4] = 9 , F[5] = 15

n >= 5  F[n] = F[n-1]+F[n-3]+F[n-4]

得到的矩阵是：

  0 1 0 0

  0 0 1 0

  0 0 0 1

  1 1 0 1

CODE：

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <string>#include <stack>#include <queue>#include <vector>#include <set>#include <map>typedef long long ll;const int inf=0xffffff;using namespace std;typedef vector<int> vec;typedef vector<vec> mat;int K,M;mat mul(mat &A,mat &B){    mat C(A.size(),vec(B[0].size()));    for(int i=0;i<A.size();i++)    {        for(int k=0;k<B.size();k++)        {            for(int j=0;j<B[0].size();j++)            {                C[i][j] = (C[i][j] + A[i][k]*B[k][j])%M;            }        }    }    return C;}mat pow(mat A){    mat B(A.size(),vec(A.size()));    for(int i=0;i<4;i++)        B[i][i]=1;    int n=K-4;    while(n > 0)    {        if(n & 1) B = mul(B, A);        A = mul(A, A);        n >>= 1;    }    return B;}int main(){    //freopen("in.in","r",stdin);    int f[5];    f[0]=2;f[1]=4;f[2]=6;f[3]=9;    while(~scanf("%d %d",&K,&M))    {        mat A(4,vec(4));        A[0][1]=1;A[1][2]=1;A[2][3]=1;        A[3][0]=1;A[3][1]=1;A[3][3]=1;        A = pow(A);        ll ans=0;        for(int i=0;i<4;i++)        {            ans +=(ll) (A[3][i]*f[i])%M;        }        if(K <= 4) printf("%d\n",f[K-1]%M);        else printf("%I64d\n",ans%M);    }    return 0;}