分形(哈尔滨理工大学校赛)

如今回想上学期犯了很多错误，如今再次发这篇博客就是因为这些错误。

所以这学期一定要把遇到的问题都解决，不要把问题都留到已经忘了，同时也要时常看看曾经做过的题，复习做过的，继续做曾经没做的。

然后就是这道题了。帮助我理解分形的一道题。注释和想说的全在代码里了。

Help C5

Time Limit: 1000 MS

Memory Limit: 65535 K

Total Submit: 51(9 users)

Total Accepted: 12(8 users)

Rating:

Special Judge: No

Description

Hello, I’m Sea5, and you can call me C5 instead. I want a program which can sign my name automatically. And my brothers, C0, C1, C2, C3, C4, C6, C7, C8, each of them wants one as well. Can you help us?

Input

First line is the number of test cases T(T<=8).

T lines follow, each line includes an integer N(N<=7), and you should help C(N) to sign his name.

Output

C0’s signature is ‘C’.

When you draw C(N)’s name, you should print the name using C(N-1)’s name as its element, and using the following format to draw it.

*XX

X**

*XX

(X is the element, * is blank space)

And please don’t print extra spaces at the end of line.

For example, C1’s name should be

*CC *CC

C C**

*CC But not *CC

(I use * to show you where are spaces.)

Sample Input

3
0
1
2

Sample Output

C
CC
C
CC
CC CC
C C
CC CC
CC
C
CC
CC CC
C C
CC CC

  1 ///分形打印的过程中点的辐射成多个点的相加过程是与实际相反的，
  2 ///先加的是正常最后一次辐射的大小，然后最后相加的是第一次辐射的大小，
  3 ///由于最后加法的交换行只要最后结果，
  4 ///所以最后是一样的，但如果保留过程就会错误
  5 ///这题是一道很好的模板。源自哈尔滨理工大学校赛
  6 #include<cstdio>   //分形问题递归打印解决
  7 
  8 #include<cstring>
  9 
 10 #include<iostream>
 11 #include<windows.h>
 12 using namespace std;
 13 
 14 
 15 
 16 char a[3000][3000];  //注意数组的大小   不然存不下图形  会RE
 17 
 18 int mypow(int d)
 19 
 20 {
 21 
 22     int ans = 1;
 23 
 24     for (int i = 1; i <= d; i++)
 25 
 26         ans *= 3;
 27 
 28     return ans;
 29 
 30 }
 31 
 32 void dfs(int cur, int x, int y)
 33 
 34 {
 35 
 36     if (cur == 1)
 37 
 38     {
 39 
 40         a[x][y] = 'C';
 41 
 42         return ;
 43 
 44     }
 45 
 46     int s = mypow(cur - 2);
 47 
 48     dfs(cur - 1, x + s, y); //这些按照格式来就好
 49     dfs(cur - 1, x, y + 2 * s);
 50 
 51     dfs(cur - 1, x, y + s);
 52 
 53     dfs(cur - 1, x + 2 * s, y + s);
 54     dfs(cur - 1, x + 2 * s, y + 2 * s);
 55 
 56 }
 57 
 58 int main(void)
 59 
 60 {
 61 
 62     int t;
 63 
 64     scanf("%d", &t);
 65 
 66     while (t--)
 67 
 68     {
 69 
 70         int n;
 71 
 72         scanf("%d", &n);
 73 
 74         n++;
 75 
 76         memset(a, ' ', sizeof(a));
 77 
 78         dfs(n, 1, 1);
 79 
 80         int s = mypow(n - 1);///此为最大边界
 81 
 82         for (int i = 0; i <= s; i++) //先把最大的边界找到
 83 
 84             a[i][s + 1] = '';
 85 ///次下为找到边界的方法，可以当做模板
 86         for (int i = 0; i <= s; i++) //其实格式可以每行从后往前来把每个编程遇到不是空格就停止
 87 
 88         {
 89 
 90             for (int j = s + 1; j >= 0; j--)
 91 
 92             {
 93 
 94                 if (a[i][j] == 'C') {
 95                     a[i][j + 1] = '';
 96                     break;
 97                 }
 98 
 99             }
100 
101         }
102 
103         for (int i = 1; i <= s; i++)
104 
105             printf("%s
", a[i] + 1); //a[i]+1 就是从a[1]开始输出
106 
107     }
108 
109     return 0;
110 
111 }

View Code