算法|Fractal解题笔记回归|c++

题面 A fractal is an object or quantity that displays self-similarity, in a somewhat technical sense, on all scales. The object need not exhibit exactly the same structure at all scales, but the same “type” of structures must appear on all scales.
A box fractal is defined as below :
A box fractal of degree 1 is simply
X
A box fractal of degree 2 is
X X
X
X X
If using B(n - 1) to represent the box fractal of degree n - 1, then a box fractal of degree n is defined recursively as following
B(n - 1) B(n - 1)

B(n - 1)

B(n - 1) B(n - 1)
Your task is to draw a box fractal of degree n.
Input The input consists of several test cases. Each line of the input contains a positive integer n which is no greater than 7. The last line of input is a negative integer ?1 indicating the end of input.
Output For each test case, output the box fractal using the ‘X’ notation. Please notice that ‘X’ is an uppercase letter. Print a line with only a single dash after each test case.
Sample

1 2 3 4 -1

X - X X X X X - X XX X XX X XX X X X X X X X XX X XX X XX X - X XX XX XX X XXXX X XX XX XX X X XX X XX X XX X X XX XX XX X XXXX X XX XX XX X X XX X XX X XX X X X X X X X XX X XX X XX X X XX XX XX X XXXX X XX XX XX X X XX X XX X XX X X XX XX XX X XXXX X XX XX XX X -

思路 【算法|Fractal解题笔记】这道题显然是个递归问题，需要我们找到递归方程，不难发现分形X的相对位置是确定的，我们可以由此推出每一层位置的递归关系；当然另一个角度看，也可以看作图像的平移。为了数组处理方便，我们使用左上角为基点就可以得到一组递归方程，如果不用左上角为基点，就要涉及原点坐标的平移变换，相对麻烦些。
AC代码

#include #include #include #include using namespace std; int n, W, cnt; char s[1001][1001]; void b(int d, int x, int y){ if(d==1)s[y][x]='X'; else { int size = 1; for(int i=2; i0; j--) if(s[i][j-1]=='X'){ s[i][j]='\0'; break; } printf("%s\n",s[i]); } printf("-\n"); scanf("%d", &n); }while(n != -1); }