Quad Tiling
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5008 Accepted: 2269

Description

Tired of the Tri Tiling game finally, Michael turns to a more challengeable game, Quad Tiling:

In how many ways can you tile a 4 × N (1 ≤ N ≤ 109) rectangle with 2 × 1 dominoes? For the answer would be very big, output the answer modulo M (0 < M ≤ 105).

Input

Input consists of several test cases followed by a line containing double 0. Each test case consists of two integers, N and M, respectively.

Output

For each test case, output the answer modules M.

Sample Input
1 10000
3 10000
5 10000
0 0

Sample Output
1
11
95

Source

POJ Monthly--2007.10.06, Dagger

问题链接：POJ3420 Quad Tiling
问题简述：（略）
问题分析：
    递推式如下：
a[0]=1，a[1]=1，a[2]=5，a[3]=11，
a[n]=a[n-1]+5a[n-2]+a[n-3]-an-4
    采用矩阵快速模幂计算，模板题。关键是找到那个递推式。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* POJ3420 Quad Tiling */

#include<cstdio>
#include<cstring>

using namespace std;

const int N = 10;
const int M = 10;

int r, mod;

struct Matrix
{
    int n,m;
    int a[N][M];
    void clear()
    {
        n = m = 0;
        memset(a, 0, sizeof(a));
    }
    Matrix operator +(const Matrix &b) const
    {
        Matrix c;
        c.n = n;
        c.m = m;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                c.a[i][j] = a[i][j] + b.a[i][j];
        return c;
    }
    Matrix operator -(const Matrix &b) const
    {
        Matrix c;
        c.n = n;
        c.m = m;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                c.a[i][j] = a[i][j] - b.a[i][j];
        return c;
    }
    Matrix operator *(const Matrix &b) const
    {
        Matrix c;
        c.clear();
        c.n = n;
        c.m = b.m;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < b.m; j++)
                for(int k = 0; k < m; k++)
                {
                    c.a[i][j] += a[i][k] * b.a[k][j];
                    c.a[i][j] %= mod;
                }
        return c;
    }
    Matrix powermod(int x)
    {
        Matrix c;
        c.clear();
        c.n = c.m = n;
        for(int i = 0; i < n; i++)
            c.a[i][i] = 1;
        if(x == 0)
            return c;
        else if(x == 1)
            return *this;

        Matrix d = powermod(x / 2);
        d = d * d;
        if(x % 2 )
            d = d * (*this);
        return d;
    }
};

int main()
{
    while(scanf("%d%d", &r, &mod) == 2 && (r || mod)) {
        int a[] = {1, 1, 5, 11};
        if(r <= 3) {
            printf("%d\n", a[r] % mod);
            continue;
        }
        Matrix b;
        b.clear();
        b.n = b.m = 4;
        b.a[0][1] = b.a[1][2] = b.a[2][3] = 1;
        b.a[3][0] = -1;
        b.a[3][1] = 1;
        b.a[3][2] = 5;
        b.a[3][3] = 1;
        b = b.powermod(r - 3);

        Matrix x;
        x.clear();
        x.n = 4;
        x.m = 1;
        x.a[0][0] = 1;
        x.a[1][0] = 1;
        x.a[2][0] = 5;
        x.a[3][0] = 11;
        x = b * x;
        printf("%d\n",(x.a[3][0] + mod) % mod);
    }

    return 0;
}