Description
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 Input
1
2
3
4
-1Sample Output
X
-
X X
X
X X
-
X X X X
X X
X X X X
X X
X
X X
X X X X
X X
X X X X
-
X X X X X X X X
X X X X
X X X X X X X X
X X X X
X X
X X X X
X X X X X X X X
X X X X
X X X X X X X X
X X X X
X X
X X X X
X X
X
X X
X X X X
X X
X X X X
X X X X X X X X
X X X X
X X X X X X X X
X X X X
X X
X X X X
X X X X X X X X
X X X X
X X X X X X X X
一递归:
先看下较为常规的递归解决:
由于图形是重复的,小的图形只是把大图形的左上角一部分输出,,只计算最大的图形,打表可加快速度。
#include#include int p[8] = {1,3,9,27,81,243,729}; char map[730][730]; //n当前的图形大小,x,y图形所在的坐标 void print(int n,int x,int y){ if(n == 0){ map[x][y] = 'X'; // return; } print(n-1, x, y); //左上 print(n-1, x+2*p[n-1], y); //右上 print(n-1, x+p[n-1], y+p[n-1]); //中间 print(n-1, x, y+2*p[n-1]); print(n-1, x+2*p[n-1], y+2*p[n-1]); } int n; int main(){ for(int i=0; i = 0){ for(int i=0; i
二 数学方法
在discuss里面看到这个短小精悍的程序。
#include"stdio.h" #include"math.h" main() { int i,j,n,ii,jj,k; while(scanf("%d",&n)&&n--!=-1) { for(i=0;i关于这个图形,还可以用来证明
寻找1/5 + 1/25 + 1/125 + .. = 1/4的图形证明
