Recursive algorithm--box FRACTAL box fractals (POJ2083)

Problem

The box fractal is defined as follows:
The 1-degree box fractals are:
X
The 2-degree box fractals are:
x x
X
x x

If B (n-1) represents the box fractal of the n-1 degree, then the box fractal recursion of the n degree is defined as follows:

B (n-1) b (n-1)
B (n-1)
B (n-1) b (n-1)

Please draw an N-degree box fractal shape

Input

Each row gives a positive integer that is not greater than 7. The last line entered ends with-1 for the input

Output

For each test case, a box fractal marked with ' X ' is written out. After each test case, the output contains a line with a dash "-".

Analysis

The size of the box fractal of n degrees is 3^ (n-1), that is, the box fractal of n degrees is a square with a width of 3^ (n-1).
Set the recursive function Printbox (n,x,y) to generate an N-degree box fractal with coordinates (x, y) as the upper-left corner.

1) Recursive boundary: If n=1, Output ' x ' at (x, y)
2) If n>1, the size of the box to calculate n-1 degrees M = 3^ (n-2), respectively, in the upper left, upper right, middle, lower left and bottom right to draw 5 n-1 degrees of box.
For the N-1 box at the top left, the coordinates of the upper-left corner are (x, y), and the recursive Printbox (n-1,x,y) is generated;
For the box with the N-1 degree in the upper right, the coordinates of the upper left corner are (x+2m,y), the recursive Printbox (n-1,x+2m,y) is generated;
For the Middle n-1 box, the coordinates of the upper left corner are (x+m,y+m), the recursive Printbox (n-1,x+m,y+m) is generated;
For the lower left n-1 box, the upper-left corner coordinates are (x,y+2m), and the recursive Printbox (n-1,x,y+2m) is generated;
For the box with n-1 degree in the upper right, the coordinates of the upper left corner are (x+2m,y+2m), the recursive Printbox (n-1,x+2m,y+2m) is generated;

Encoding implementation

#include "stdafx.h"#include <cmath>#include <iostream>using namespace STD;//7-degree box fractal largest n=3^6=729#define MAX 730CharMaps[max][max];voidPrintbox (intNintXintY) {//Recursive boundary    if(n = =1) {Maps[x][y] =' X '; }Else{size of//n-1 box fractal m        intm =POW(3N2);///n-1 box Fractal at upper leftPrintbox (N-1, x, y);///n-1 box fractals on top rightPrintbox (n1, x+2*m, y);//N-1 in the middle of the box fractalPrintbox (N-1, X, Y +2* m);///n-1-box fractals at lower leftPrintbox (N-1, X + M, y + m);//n-1-box fractals at lower rightPrintbox (n1, x+2*m,y+2*M); }}int_tmain (intARGC, _tchar* argv[]) {intn;Cin>> N; while(N! =-1){intSize =POW(3N1);//Initialize         for(inti =0; i < size; i++) { for(intj =0; J < size; J + +) {Maps[i][j] ="'; Maps[i][size] =' + '; }} printbox (N,0,0);//Output         for(inti =0; i < size; i++)printf("%s\n", Maps[i]);cout<<"-"<<endl;Cin>> N; }return 0;}

Test

Recursive algorithm--box FRACTAL box fractals (POJ2083)