Sicily 1527. tiling a grid with dominoes

There is a 4xn Board, an unlimited 1X2 Domino card. Enter n to calculate the number of methods that cover the Board perfectly.

The method is as follows:

The two-dimensional array DP [N] [m] indicates the number of arrangement methods when n columns exist and the column status is M. M is represented by four bits in binary format. The a bit indicates the status of row A of the column. 1 indicates that the column is overwritten, and 0 indicates that the column is not overwritten. The initial status is:

DP [1] [3] = DP [1] [6] = DP [1] [12] = DP [1] [15] = DP [1] [0] = 1;

Status transfer:

DP [T] [15] = DP [T-1] [15] + dp [T-1] [12] + dp [T-1] [3] + dp [T-1] [6] + DP [T-1] [0];
DP [T] [9] = DP [T-1] [6];
DP [T] [12] = DP [T-1] [15] + dp [T-1] [3];
DP [T] [3] = DP [T-1] [15] + dp [T-1] [12];
DP [T] [6] = DP [T-1] [15] + dp [T-1] [9];
DP [T] [0] = DP [T-1] [15];

Code:

// Problem#: 8223// Submission#: 2127308// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <iostream>#include <algorithm>#include <cstring>using namespace std ; int main(){    int d[31][16] ;    memset(d,0,sizeof(d)) ;     d[1][15] =  d[1][12] = d[1][6] =  d[1][3] =d[1][0]= 1 ;     for(int i = 2 ; i <= 30 ; i ++)    {        d[i][15] = d[i-1][0]+d[i-1][3]+d[i-1][6]+d[i-1][12]+d[i-1][15] ;        d[i][12] = d[i-1][3] + d[i-1][15] ;        d[i][9] = d[i-1][6] ;        d[i][6] = d[i-1][15] + d[i-1][9] ;        d[i][3] = d[i-1][15] + d[i-1][12] ;        d[i][0] = d[i-1][15] ;    }    int k = 1 ;     int n ;cin >> n ;    while(n--)    {        int c ;        cin >> c ;        cout << k++ <<" " << d[c][15] << endl;     }}