POJ3420Quad Tiling（矩阵快速幂）

最新推荐文章于 2021-09-11 19:55:36 发布

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原文链接：http://www.cnblogs.com/juechen/p/5255898.html

本文探讨了使用矩阵快速幂优化解决四边形瓷砖铺砌问题的方法，具体涉及到矩阵乘法、快速幂算法以及模运算的应用。

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Quad Tiling

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 3740 Accepted: 1684
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

递推式：a[i]=a[i-1]+5*a[i-2]+a[i-3]-a[i-4];

由于N高达10^9，所以要用矩阵进行优化。
|0 1 0 0|
|0 0 1 0|
|0 0 0 1|
|-1 1 5 1|
与
|a[i-3]|
|a[i-2]|
|a[i-1]|
|a[i]|
相乘

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <cstdlib>
#include <algorithm>
#define LL long long

using namespace std;
const int  Max = 10;

int Mod;
struct Matrix
{
    int n,m;
    int a[Max][Max];
    void clear()//清空矩阵
    {
        n=0;
        m=0;
        memset(a,0,sizeof(a));
    }
    Matrix operator * (const Matrix &b)const//矩阵相乘
    {
        Matrix tmp;
        tmp.clear();
        tmp.n=n;
        tmp.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++)
                {
                    tmp.a[i][j]=(tmp.a[i][j]+(a[i][k]%Mod)*(b.a[k][j]%Mod))%Mod;
                }
            }
        }
        return tmp;
    }
};



void Pow(int m)
{
    Matrix s;
    s.clear();
    s.n=4;
    s.m=4;
    s.a[3][3]=1;s.a[3][2]=5;
    s.a[3][1]=1;s.a[3][0]=-1;
    s.a[1][2]=1;s.a[2][3]=1;
    s.a[0][1]=1;

    Matrix ans;
    ans.clear();
    ans.n=4;
    ans.m=1;
    ans.a[0][0]=1;
    ans.a[1][0]=5;
    ans.a[2][0]=11;
    ans.a[3][0]=36;
    while(m)//快速幂
    {
        if(m&1)
        {
            ans=s*ans;
        }
        s=s*s;
        m>>=1;
    }
    printf("%d\n",ans.a[3][0]);
}

int main()
{ 
    int n;
    while(scanf("%d %d",&n,&Mod),n)
    {
        if(n<4)
        {
            switch(n)
            {
                case 1:
                    printf("%d\n",1%Mod);
                    break;
                case 2:
                    printf("%d\n",5%Mod);
                    break;
                case 3:
                    printf("%d\n",11%Mod);
                    break;
            }
            continue;
        }
        Pow(n-4);
    }
    return 0;
}