Queuing（矩阵快速幂）

Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2586    Accepted Submission(s): 1210

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

     题意：

     给出 L 和 M，代表 L 长度的队伍，f 代表女生，m 代表有男生，存在 fmf 或者 fff 的队伍叫 O - queue，没有叫做 E - queue，求出 L 长度序列的队伍有多少种 E - queue 的可能，结果 % M。

 

     思路：

     矩阵快速幂。

     设 ai 为以 fm 为后缀的包含序列的种数；

          bi 为以 ff 为后缀的包含序列的种数；

             ci 为以 mm 为后缀的包含序列的种数；

         di 为以 mf 为后缀的包含序列的种数；

         ei 为以 fm 为后缀的不包含序列的种数；

         fi 为以 ff 为后缀的不包含序列的种数；

         gi 为以 mm 为后缀的不包含序列的种数；

         hi 为以 mf 为后缀的不包含序列的种数。

     所以：

     ai+1 = bi + ci； bi+1 = bi + ci + di ； ci+1 = ai + di + ei ； di+1 = ai + di；

     ei+1 = fi + gi； fi+1 = gi ；gi+1 = hi ； hi = ei + hi

     所以可以得出矩阵：

     
 

     

     AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

typedef vector<int> vec;
typedef vector<vec> mat;

mat mul (mat a, mat b, int mod) {
    mat c(a.size(), vec(b[0].size()));

    for (int i = 0; i < a.size(); ++i) {
        for (int j = 0; j < b[0].size(); ++j) {
            for (int k = 0; k < b.size(); ++k) {
                c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % mod;
            }
        }
    }

    return c;
}

mat pow (mat a, int n, int mod) {
    mat b(a.size(), vec(a[0].size()));
    for (int i = 0; i < a.size(); ++i) {
        b[i][i] = 1;
    }

    while(n > 0) {
        if (n & 1) b = mul(b, a, mod);
        a = mul(a, a, mod);
        n >>= 1;
    }

    return b;
}

int main() {

    int l, m;
    while (~scanf("%d%d", &l, &m)) {

        mat a(8, vec(8));
        a[0][1] = 1, a[0][2] = 1, a[1][1] = 1;
        a[1][2] = 1, a[1][5] = 1, a[2][0] = 1;
        a[2][3] = 1, a[2][4] = 1, a[3][0] = 1;
        a[3][3] = 1, a[4][5] = 1, a[4][6] = 1;
        a[5][6] = 1; a[6][7] = 1, a[7][4] = 1;
        a[7][7] = 1;

        mat b(8, vec(1));
        b[1][0] = 1, b[2][0] = 1, b[4][0] = 2;
        b[5][0] = 1, b[6][0] = 1, b[7][0] = 2;

        if (l > 3) {

            a = pow(a, l - 3, m);
            b = mul(a, b, m);

            int sum = 0;
            for (int i = 4; i < 8; ++i) {
                sum = (sum + b[i][0]) % m;
            }
            printf("%d\n", sum % m);
            
        } else if (l == 3) printf("%d\n", 6 % m);
        else printf("0\n");

    }

    return 0;
}