Fractal   Problem 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. be careful, no extra blankspace allowed in the output.  Sample Input 12-1 Sample Output X-X X  XX X-X X   X X X     XX X   X X   X X    X   X XX X   X X X     XX X   X X // 提交了几次，开始是超时，后来改进如下// 可能不超时，但是Wrong Answer，测试怎么都没测出来// 很多情况都考虑到了包括空格，回车，最后终于发现用管道测试时// 如果输入文件里只有7（-1也在里面）得到的结果是乱码// 不知道是不是错在这里。就算错了也不知道怎么改，我尝试了加大数组// 甚至改为C++代码，结果仍是乱码，极度郁闷 ！ #include <stdio.h>#include <string.h>#include <stdlib.h> int       a[] = {1,3,9,27,81,243,729}; char   metrix[800][800]; void set(int x,int y,int n){     metrix[x][y] = 'X';     metrix[x-1][y-1] = 'X';     metrix[x+1][y-1] = 'X';     metrix[x+1][y+1] = 'X';     metrix[x-1][y+1] = 'X';     if(n == 2)         return ;     set(x, y, n - 1);     set(x - a[n-2],y - a[n-2],n - 1);     set(x + a[n-2],y - a[n-2],n - 1);     set(x + a[n-2],y + a[n-2],n - 1);     set(x - a[n-2],y + a[n-2],n - 1);} int main(){    int  n,i,j,flag = 0;    char saveIndex[800];     memset(metrix,' ',800*800);    set(364,364,7);    while(1){        scanf("%d",&n);        if(n == -1)            break;        if(flag)            printf("\n");        if(n == 1){            printf("X\n-");            flag = 1;            continue;        }        for(i = 0; i < a[n-1]; i++){            for(j = a[n-1]; 1; j --){                if(metrix[i][j] != ' ')                    break;            }            saveIndex[i] = j+1;            metrix[i][j+1] = '\0';        }            for(i = 0; i < a[n-1]; i++)            printf("%s\n",metrix[i]);        for(i = 0; i < a[n-1]; i++)            metrix[i][saveIndex[i]] = ' ';        printf("-");        flag = 1;    }    return 0;}

评论