hdoj 2604 Queuing

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Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6110 Accepted Submission(s): 2661

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 2L 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

题目解析：

这道题目的关键在于对这个队列的结尾进行建模。从L=2 开始进行讨论，如果要是E队列，若原先结尾为ff,可以变化为fm，而原先结尾为fm, 可以变化为mm，原先结尾为mf, 可以变化为ff和fm，原先结尾为mm，可以变化为mf和mm。因此，可以构建一个状态转移矩阵。而初始状态L=2的时候，ff = fm = mf = mm = 1个，因此，通过快速幂的方法，可以得到任意（L>2）时候的结尾为ff，fm，mf和mm的序列的个数，最后求和即可。

复制代码

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
const int INF = 2e9;
const int MAXN = 1e6;
#define REP(i, s, k) for(int i = s; i< k; i++)

int MOD;
struct matrix{
 int n;
 vector<vector<int> > M;
 matrix(int n){
  this->n = n;
  M.resize(n, vector<int>(n, 0)); 
 }
 void eye(){for(int i = 0; i < n; i++) M[i][i] = 1;}
 
 friend matrix operator *(matrix A, matrix B){
  int n = A.n; matrix res(n);
  REP(i, 0, n) REP(j,0,n){
    if(A.M[i][j] == 0) continue;
    REP(k, 0, n) res.M[i][k] = (res.M[i][k] + A.M[i][j] * B.M[j][k] % MOD) % MOD;
  }
  return res;
 }

};

inline matrix qp(matrix A, int p){
  matrix res(4); res.eye();
  for(; p; p >>=1, A=A*A){
    if(p&1) res = res * A;
  } 
  return res;
}

inline int qp(int A, int p){
  int res = 1;
  for(; p; p >>=1, A = (A * A) % MOD){
    if(p & 1) res  = (res + A) % MOD;
  } 
  return res;
}

signed main(){
  int L, M;
  while(~scanf("%d %d", &L, &M)){
    MOD = M;
    if(L <= 2) printf("%dn", pow(2, L));
    matrix m(4); m.M[0][1] = m.M[1][3] = m.M[2][0] = m.M[2][1] = m.M[3][2] = m.M[3][3] = 1;
    m = qp(m, L-2);  
    int ans = 0;
    for(auto v: m.M) for(auto e: v) ans += e;
    ans %= MOD;
    printf("%dn", ans);
  }
  return 0;
}

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#include <bits/stdc++.h>

using namespace std;
using ll = long long;
const int INF = 2e9;
const int MAXN = 1e6;
#define REP(i, s, k) for(int i = s; i< k; i++)

int MOD;
struct matrix{
 int n;
 vector<vector<int> > M;
 matrix(int n){
  this->n = n;
  M.resize(n, vector<int>(n, 0)); 
 }
 void eye(){for(int i = 0; i < n; i++) M[i][i] = 1;}
 
 friend matrix operator *(matrix A, matrix B){
  int n = A.n; matrix res(n);
  REP(i, 0, n) REP(j,0,n){
    if(A.M[i][j] == 0) continue;
    REP(k, 0, n) res.M[i][k] = (res.M[i][k] + A.M[i][j] * B.M[j][k] % MOD) % MOD;
  }
  return res;
 }

};

inline matrix qp(matrix A, int p){
  matrix res(4); res.eye();
  for(; p; p >>=1, A=A*A){
    if(p&1) res = res * A;
  } 
  return res;
}

inline int qp(int A, int p){
  int res = 1;
  for(; p; p >>=1, A = (A * A) % MOD){
    if(p & 1) res  = (res + A) % MOD;
  } 
  return res;
}

signed main(){
  int L, M;
  while(~scanf("%d %d", &L, &M)){
    MOD = M;
    if(L <= 2) printf("%dn", pow(2, L));
    matrix m(4); m.M[0][1] = m.M[1][3] = m.M[2][0] = m.M[2][1] = m.M[3][2] = m.M[3][3] = 1;
    m = qp(m, L-2);  
    int ans = 0;
    for(auto v: m.M) for(auto e: v) ans += e;
    ans %= MOD;
    printf("%dn", ans);
  }
  return 0;
}