HDU 2604 Queuing

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

矩阵快速幂。
#include<stdio.h>
#include<string.h>
#include<vector>
#include<queue>
#include<functional>
using namespace std;
const int maxn = 1000005;
const int size = 4;
int n, m, ans, f[size], p[size];

struct abc
{
    int a[size][size];
    abc(){ memset(a, 0, sizeof(a)); }
    void operator =(const abc&b) { memcpy(a, b.a, sizeof(a)); }
    abc operator *(const abc&b) {
        abc c;
        for (int i = 0; i < size;i++)
            for (int j = 0; j < size;j++)
                for (int k = 0; k < size; k++)
                    c.a[i][k] = (c.a[i][k] + a[i][j] * b.a[j][k]) % m;
        return c;
    }
};

abc get(int x)
{
    abc a, b, c;
    a.a[0][1] = a.a[1][3] = a.a[2][0] = a.a[2][1] = a.a[3][2] = a.a[3][3] = 1;
    for (int i = 1, f = 0; x; x >>= 1)
    {
        if (x & 1) 
            if (f) b = b * a; else b = a, f = 1;
        a = a * a;
    }
    return b;
}

int solve()
{
    int tot = 0;
    abc a = get(n - 2);
    memset(p, 0, sizeof(p));
    f[0] = f[1] = f[2] = f[3] = 1;
    for (int j = 0; j < size; j++)
        for (int k = 0; k < size; k++)
            p[k] = (p[k] + f[j] * a.a[j][k]) % m;
    for (int j = 0; j < size; j++) tot = (tot + p[j]) % m;
    return tot;
}

int main()
{
    while (scanf("%d%d", &n, &m) == 2)
    {
        if (n <= 2){ ans = 2 * n % m; }
        else ans = solve();
        printf("%d\n", ans);
    }
    return 0;
}