矩阵快速幂2 3*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;

//U[n]表示3×N棋盘的摆放数，
//
//V[n]表示缺了1×1边角的3×N棋盘的摆放数
//
//则
//
//U[n] = 2*V[n-1] + U[n-2]
//
//V[n] = U[n-1] + V[n-2]
//
//其中，
//
//U[0] = 1, U[2] = 3;
//
//V[0] = 1, V[1] =1;

//[U(n)                       [ 0 2 1 0           [ U(n-1)
// V(n)                         1 0 0 1             V(n-1)
// U(n-1)        =              1 0 0 0      *      U(n-2)
// V(n-1)]                      0 1 0 0 ]           V(n-2) ]
//
//                   [ 0 2 1 0                    [ U(1)
//               =     1 0 0 1                      V(1)
//                     1 0 0 0     ^  (n-1)         U(0)
//                     0 1 0 0 ]                    V(0) ]

const int MOD = 12357;

struct Martix
{
    LL martix[5][5];
    int row,col;
    Martix(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])%MOD;
        return C;
    }
};


void solve()
{
    Martix A(4,4), F(4,4), S(4,1);
    A.martix[0][2] = A.martix[1][0] = A.martix[1][3] = A.martix[2][0] = A.martix[3][1] = 1;
    F.martix[0][0] = F.martix[1][1] = F.martix[2][2] = F.martix[3][3] = 1;
    A.martix[0][1] = 2;
    S.martix[1][0] = S.martix[2][0] = S.martix[3][0] = 1;
    int n;
    scanf("%d",&n);
    if(n&1)
    {
        printf("%d
",0);
        return ;
    }
    n--;
    while(n > 0)
    {
        if(n & 1)
            F = F*A;
        A = A*A;
        n >>= 1;
    }
    S = F*S;
    printf("%lld
",S.martix[0][0]);
}

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