[POJ 3420] Quad Tiling

 

Quad Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3495		Accepted: 1539

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

 

川大校赛的原题出处，，，醉了，，比赛的时候一直没推出公式，唉，弱得不行

重点在求递推公式，再矩阵快速幂即可。

SCU 4430 把输入改一下就可以了

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

#define ll long long

#define N 10



int MOD;

struct Matric

{

    int size;

    int a[N][N];

    Matric(int s=0)

    {

        size=s;

        memset(a,0,sizeof(a));

    }

    Matric operator * (const Matric &t)

    {

        Matric res=Matric(size);

        for(int i=0;i<size;i++)

        {

            for(int k=0;k<size;k++)

            {

                if((*this).a[i][k])

                    for(int j=0;j<size;j++)

                    {

                        res.a[i][j]+=(ll)(*this).a[i][k]*t.a[k][j]%MOD;

                                    res.a[i][j]=(res.a[i][j]+MOD)%MOD;

                    }

            }

        }

        return res;

    }

    Matric operator ^ (int n)

    {

        Matric ans=Matric(size);

        for(int i=0;i<size;i++) ans.a[i][i]=1;

        while(n)

        {

            if(n&1) ans=ans*(*this);

            (*this)=(*this)*(*this);

            n>>=1;

        }

        return ans;

    }

    void debug()

    {

        for(int i=0;i<size;i++)

        {

            for(int j=0;j<size;j++)

            {

                printf("%d ",a[i][j]);

            }

            printf("\n");

        }

    }

};

int main()

{

    int n;

    while(scanf("%d%d",&n,&MOD),n||MOD)

    {

        Matric a=Matric(4);

        Matric b=Matric(4);

        a.a[0][0]=1;

        a.a[0][1]=1;

        a.a[0][2]=5;

        a.a[0][3]=11;



        b.a[0][3]=-1;

        b.a[2][3]=5;

        b.a[1][0]=b.a[2][1]=b.a[3][2]=b.a[1][3]=b.a[3][3]=1;



        b=b^n;

        a=a*b;

        printf("%d\n",(a.a[0][0]+MOD)%MOD);

    }

    return 0;

}