[POJ 3420] Quad Tiling

Quad Tiling

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;
}