HDU 2604 Queuing

最新推荐文章于 2020-07-08 19:22:33 发布

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#ACM新手上路 #c++ #矩阵快速幂 #dp #HDU

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本文介绍了一个算法，用于计算给定长度的队列中偶数序列的数量，考虑到队列中元素的性别标识。通过动态规划和矩阵快速幂，解决了这一问题，并提供了AC代码实现。

Queuing

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

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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

Sample Output

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;
}