HDU 2604 Queuing

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我是靠谱客的博主威武河马，这篇文章主要介绍HDU 2604 Queuing，现在分享给大家，希望可以做个参考。

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

复制代码

Sample Output

复制代码

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("%dn", ans);
    }
    return 0;
}

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6
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矩阵快速幂。
复制代码#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("%dn", ans);
    }
    return 0;
}#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("%dn", ans);
    }
    return 0;
}