Board coverage issues (C + + implementation)

in a chessboard of 2kx2k squares, a square is different from other squares, calling it a special square and calling it a special chessboard. problem: Use 4 different forms of L-shaped dominoes to cover all squares except special squares on a given special chessboard, and any 2 must not overlap. There are 4k cases where special squares appear on the board. Thus, for any k>=0, there are 4k different special chessboard. It is easy to know that in any 2k * 2k chessboard, the number of L-type dominoes used is exactly (4k-1)/3.

1. When k>0, divide the 2kx2k checkerboard into 4 2k-1x2k-1 sub-chessboard, as shown in figure (a).
2. Special squares must be located in one of the 4 smaller sub-chessboard, with no special squares in the remaining 3 sub-chessboard.
3. To convert the non-special square checkerboard to a special chessboard, a domino can be used to cover the 3 smaller checkerboard's rendezvous, as shown in figure (b), thus translating the original problem into 4 smaller checkerboard coverage issues.
4. Use this division recursively until the chessboard is simplified to a checkerboard 1x1.

#include <iostream>using namespacestd;inttile=1;//number of L-type dominoes (increment)intboard[ -][ -];//Chess Board/****************************************************** Recursive way to implement the board overlay algorithm * Input parameters: * tr--The line number in the upper left corner of the current checkerboard * tc--the column number in the upper left corner of the current checkerboard * dr--the line number of the current special box * dc--the column number where the current special square is located * Size: The current chessboard: 2^k*****************************************************/voidChessboard (intTrintTcintDrintdcintsize) {    if(size==1)return; intt=tile++,s=size/2; if(Dr<tr+s && Dc<tc+s)///in the upper left corner areachessboard (tr,tc,dr,dc,s); Else///not in the upper-left corner{board[tr+s-1][tc+s-1]=t;///with the t number (denoted by a number) L-type dominoes cover the lower right cornerChessboard (tr,tc,tr+s-1, tc+s-1, s);///cover the remaining squares    }    if(Dr<tr+s && Dc>=tc+s)///in the upper right corner areaChessboard (tr,tc+s,dr,dc,s); Else///not in the upper-right corner of the area{board[tr+s-1][tc+s]=T; Chessboard (TR,TC+s,tr+s-1, tc+s,s); }    if(Dr>=tr+s && dc<tc+s) Chessboard (tr+s,tc,dr,dc,s); Else{board[tr+s][tc+s-1]=T; Chessboard (tr+s,tc,tr+s,tc+s-1, s); }    if(Dr>=tr+s && dc>=tc+s) Chessboard (tr+s,tc+s,dr,dc,s); Else{board[tr+s][tc+s]=T; Chessboard (tr+s,tc+s,tr+s,tc+s,s); }}intMain () {intsize; cout<<"Enter the size of the chessboard (the magnitude must be 2 n Power):"; CIN>>size; intindex_x,index_y; cout<<"enter the coordinates of the special grid position:"; CIN>>index_x>>index_y; Chessboard (0,0, index_x-1, index_y-1, size);  for(intI=0; i<size; i++ )     {          for(intj=0; j<size; J + +) cout<<Board[i][j]<<'\ t'; cout<<Endl; }}

Derivation process: The original equivalent to T (k) =4t (k-1) +1 recursion: 4T (k-1) =4 (4T (k-2) +1) =42t (k-2) +4 T (k) = 42T (k-2) +4 +1                       Also: 42T (k-2) =43t (k-3) +42 so T (k) = 43T (k-3) +42+4+1 ........... T (k) =4kt (0) +4k-1+...+4+1=o (4k)

Board coverage issues (C + + implementation)