POJ1941 The Sierpinski Fractal

 1 //It is made by jump~
 2 #include <iostream>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <algorithm>
 8 #include <ctime>
 9 #include <vector>
10 #include <queue>
11 #include <map>
12 #include <set>
13 using namespace std;
14 typedef long long LL;
15 char ch[2048][2048];
16 int n;
17 
18 inline int getint()
19 {
20        int w=0,q=0; char c=getchar();
21        while((c<‘0‘ || c>‘9‘) && c!=‘-‘) c=getchar(); if(c==‘-‘) q=1,c=getchar(); 
22        while (c>=‘0‘ && c<=‘9‘) w=w*10+c-‘0‘, c=getchar(); return q ? -w : w;
23 }
24 
25 inline void solve(int x,int y,int now){
26     if(now==1) { 
27     ch[x][y]=‘/‘; ch[x][y+1]=‘_‘; ch[x][y+2]=‘_‘; ch[x][y+3]=‘\\‘;
28     ch[x-1][y+1]=‘/‘; ch[x-1][y+2]=‘\\‘;
29     return ;
30     }
31     int du=(1<<now);//拆分成三个小三角形
32     solve(x,y,now-1); solve(x,y+du,now-1); solve(x-du/2,y+du/2,now-1);
33 }
34 
35 inline void work(){
36     while(1) {
37     n=getint(); if(n==0) break;
38     memset(ch,0,sizeof(ch)); int mi=(1<<n);
39     solve(mi,0,n); int last;//记录最后一个
40     for(int i=1;i<=mi;i++) {
41         for(int j=0;j<mi*2;j++) if(ch[i][j]) last=j;
42         for(int j=0;j<last;j++) if(!ch[i][j]) ch[i][j]=‘ ‘;
43     }
44     for(int i=1;i<=mi;i++) printf("%s\n",ch[i]);
45     printf("\n");
46     }
47 }
48 
49 int main()
50 {
51   work();
52   return 0;
53 }