哈理工OJ 2291 Help C5（分形题目）

题目链接：http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=2291
Help C5
Time Limit: 1000 MS Memory Limit: 65535 K
Total Submit: 170(39 users) Total Accepted: 50(37 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

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上面的图形和原题中不符，建议去原题中看图形------------------------------------------ -------------------------------------下面是AC代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char a[3000][3000];int mypow(int n){    int ans=1;    for(int i=1; i<=n; i++)    {        ans=ans*3;    }    return ans;}void dfs(int cur,int x,int y){    if(cur==1)    {        a[x][y]='C';        return ;    }    int s=mypow(cur-2);    dfs(cur-1,x,y+s);    dfs(cur-1,x,y+2*s);    dfs(cur-1,x+s,y);    dfs(cur-1,x+2*s,y+s);    dfs(cur-1,x+2*s,y+2*s);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n;        memset(a,' ',sizeof(a));        scanf("%d",&n);        dfs(n+1,1,1);        int s=mypow(n);        for(int i=1; i<=s; i++)        {            for(int j=s;j>=1;j--)                if(a[i][j]=='C')                {                    a[i][j+1]='\0';//去除每行里面多余的空格                    break;                }        }        for(int i=1; i<=s; i++)        {            printf("%s\n",a[i]+1);//从每行的第二个开始输出        }    }    return 0;}