HDU 2604 Queuing

Queuing

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status

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 
     

 

分析
dp
状态转移方程的推导
计算f(n)时，当第 n 位为 m 时，对前面无影响，所以记f1 = f(n-1),当第 n 位为 f 时，观察后两位，不论后两位为 mf 还是 ff 都会对前面产生限制，继续观察后三位，若后三位为 fff 和 fmf 时不满足条件，当结尾为 mmf 时对前面无影响，记 f2 = f(n-3)，当结尾为 mff 时，对前面产生限制，继续观察后四位，mmff和fmff中，只有mmff符合条件且不会对前面产生影响，记 f3 = f(n-4)。至此，所有可能的条件都已经找齐，所以 f(n) = f1 + f2 + f3，即 f(n) = f(n-1) + f(n-3) + f(n-4)。

矩阵快速幂
构造的矩阵A为 4*4 的矩阵，如下
0 1 0 0
0 0 1 0
0 0 0 1
1 1 0 1
要求 f(n)，只需乘 n-3 次即可。

AC代码如下

#include<cstdio>
#include<cstring>
#define maxn 4
using namespace std;
struct S
{
    int m[4][4];
};
int a[maxn];
int l,m;

S mul(S a,S b)
{
    S c;
    memset(c.m,0,sizeof(c.m));
    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] += a.m[i][k] * b.m[k][j];
            c.m[i][j] %= m;
        }
    return c;
}

S quick_power(S a,int k)
{
    if(k == 1) return a;
    else if(k & 1) return mul(quick_power(a,k-1),a);
    else
    {
        S c = quick_power(a,k >> 1);
        return mul(c,c);
    }
}

int main()
{
    a[0] = 1,a[1] = 2,a[2] = 4,a[3] = 6;
    S c;
    for(int i = 0 ; i < 4 ; i++)
        if(i == 2) c.m[3][i] = 0;
        else c.m[3][i] = 1;
    for(int i = 0 ; i < 3; i++)
        for(int j = 0 ; j < 4; j++)
        {
            if(j == i + 1) c.m[i][j] = 1;
            else c.m[i][j] = 0;
        }
    while(~scanf("%d%d",&l,&m))
    {
        if(l < 4)
        {
            printf("%d\n",a[l]%m);
            continue;
        }
        S b = quick_power(c,l-3);
        int ans = 0;
        for(int i = 0 ; i < 4 ; i++)
            ans += (b.m[3][i] * a[i]) % m;
        printf("%d\n",ans%m);
    }
    return 0;
}