矩阵快速幂3 k*n铺方格

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <cstdio>
#include <algorithm>
#include <map>
#include <time.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define LL long long

using namespace std;
using namespace __gnu_pbds;


const int MOD = 12357;

struct Martix
{
    LL martix[260][260];
    int row,col;
    Martix(int _row,int _col)
    {
        memset(martix,0,sizeof(martix));
        row = _row;
        col = _col;
    }
    void sets(int _row,int _col)
    {
        memset(martix,0,sizeof(martix));
        row = _row;
        col = _col;
    }
    Martix operator *(const Martix &A)const
    {
        Martix C(row, A.col);
        for(int i = 0; i < row; i++)
            for(int j = 0; j < A.col; j++)
                for(int k = 0; k < col; k++)
                {
                    C.martix[i][j] =  (C.martix[i][j] + martix[i][k] * A.martix[k][j]);
                    if(C.martix[i][j] >= MOD)
                        C.martix[i][j]%=MOD;
                }

        return C;
    }
};

//
//第i行不放置：new_x = x << 1, new_y = (y << 1) + 1; 列数+1
//第i行竖放骨牌：new_x = (x << 1) + 1, new_y = y << 1; 列数+1
//第i行横向骨牌：new x = (x << 2) + 3, new_y = (y << 2) + 3; 列数+2

int k;
Martix A(260,260),F(260,260);
void dfs(int x,int y,int col)
{
    if(col == k) {A.martix[y][x] = 1; return ;}
    dfs(x<<1, (y<<1) + 1, col+1);
    dfs( (x<<1) + 1, y << 1, col + 1);
    if(col + 2 <= k)
        dfs( (x << 2)+ 3, (y << 2)+3, col+2);
}

void solve()
{
    int n;
    scanf("%d %d",&k,&n);
    if( (k&1) && (n&1) )
    {
        printf("%d
",0);
        return ;
    }
    A.sets(1<<k,1<<k);
    F.sets(1<<k,1<<k);
    dfs(0,0,0);
    for(int i = 0; i < (1<<k); i++)
        F.martix[i][i] = 1;
    while(n > 0)
    {
        if(n & 1)
            F = F*A;
        A = A*A;
        n >>= 1;
    }
    printf("%lld
",F.martix[ (1<<k)-1 ][ (1<<k)-1 ]);
}

int main(void)
{
    solve();
    return 0;
}