hdu2604 Queuing

Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6610    Accepted Submission(s): 2903

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

 

可以看成只有0 1 串的字符串，长度为n的字符串中子串中不出现101 和111这两个的总数量有多少。然后求模m

恨容易推出a_n = a_n-1+a_n-3+a_n-4，然后m是一个不变的话，就可以直接循环遍历了。但m是个变量。

由于a_n = a_n-1+a_n-3+a_n-4 ，所以可以得到：

a_n　　　　　　1  　　0 　　1　　1　　　　a_n-1

a_n-1　　　　　 1　　  0　　 0 　  0　　　　a_n-2

an-2     =     　 0　　  1　　 0　   0     *        a_n-3

a_n-3  　　　　   0　　  0　　 1　　0　　　   a_n-4

 

得到公式就好写了。

// a[n] = a[n-1]+a[n-3]+a[n-4];
#include <bits/stdc++.h>
using namespace std;
int n, m;
struct mat{
    int m[4][4];
    mat() {
        memset(m, 0, sizeof(m));
    }
};
mat mul(mat &A, mat &B) {
    mat C;
    for(int i = 0; i < 4; i ++) {
        for(int j = 0; j < 4; j ++) {
            for(int k = 0; k < 4; k ++) {
                C.m[i][j] = (C.m[i][j] + A.m[i][k]*B.m[k][j]) % m;
            }
        }
    }
    return C;
}
mat pow(mat A, int n){
    mat B;
    for(int i = 0; i < 4; i ++) B.m[i][i] = 1;
    while(n) {
        if(n&1) B = mul(B,A);
        A = mul(A,A);
        n >>= 1;
    }
    return B;
}
int main() {
    while(cin >> n >> m) {
        mat A;
        if(n==0) printf("%d\n",1%m);
        else if(n==1) printf("%d\n",2%m);
        else if(n==2) printf("%d\n",4%m);
        else if(n==3) printf("%d\n",6%m);
        else {
            A.m[0][0] = A.m[0][2] = A.m[0][3] = 1;
            A.m[1][0] = A.m[2][1] = A.m[3][2] = 1;
            A = pow(A,n-3);
            int ans = A.m[0][0]*6+A.m[0][1]*4+A.m[0][2]*2+A.m[0][3];
            printf("%d\n",ans%m);
        }
    }
    return 0;
}