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
一遞歸：
先看下較為常規的遞歸解決：
由於圖形是重復的，小的圖形只是把大圖形的左上角一部分輸出，，只計算最大的圖形，打表可加快速度。
#include <stdio.h>
#include <string.h>
int p[8] = {1,3,9,27,81,243,729};
char map[730][730];
//n當前的圖形大小，x，y圖形所在的坐標
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]); //中間
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<p[6]; i++) memset(map[i], 32, p[6]);
print(6, 0, 0); //打表
while(scanf("%d", &n) && n-- >= 0){
for(int i=0; i<p[n]; i++){
map[i][p[n]]=0;
puts(map[i]);
map[i][p[n]]=' ';
}
puts("-");
}
return 0;
}
二 數學方法
在discuss裡面看到這個短小精悍的程序。
#include"stdio.h"
#include"math.h"
main()
{
int i,j,n,ii,jj,k;
while(scanf("%d",&n)&&n--!=-1)
{
for(i=0;i<pow(3,n);i++,printf("\n"))
for(j=0;j<pow(3,n);j++)
{
for(ii=i,jj=j,k=0;k<n&&(ii%3+jj%3)%2==0;ii/=3,jj/=3,k++);
printf("%c",32+56*(k==n));
}
printf("-\n");
}
}
關於這個圖形，還可以用來證明
尋找1/5 + 1/25 + 1/125 + .. = 1/4的圖形證明
