POJ 1244 Slots of Fun(计算几何)

很简单的一题，数据很小，直接暴力的。但是也是写也好久，有几个数，没算好。。。一次CE，一次PE，3Y。
 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <cstdlib>
 5 #include <cmath>
 6 using namespace std;
 7 #define eps 1e-8
 8 #define zero(x) (((x) > 0?(x):-(x))<eps)
 9 int o[200];
10 struct point
11 {
12     double x,y;
13 } p[1001];
14 double dis(point p1,point p2)
15 {
16     return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);
17 }
18 int judge(int i,int j,int k)
19 {
20     if(zero(dis(p[i],p[j])-dis(p[j],p[k]))&&zero(dis(p[i],p[j])-dis(p[i],p[k])))
21     return 1;
22     else
23     return 0;
24 }
25 char str[1001];
26 int main()
27 {
28     int n,i,j,k,num = 1,len,flag;
29     double d;
30     p[0].x = 0;
31     p[0].y = 0;
32     for(i = 2; i <= 12; i ++)
33     {
34         d = (i-1)*1.0*sqrt(3.0)*2.0/3.0/(i-1);
35         p[num].x = -(i-1)*1.0*sqrt(3.0)/3.0;
36         p[num].y = i-1;
37         num ++;
38         for(j = 1; j < i; j ++)
39         {
40             p[num].x = p[num-1].x + d;
41             p[num].y = i-1;
42             num ++;
43         }
44     }
45     while(scanf("%d",&n)!=EOF)
46     {
47         if(n == 0) break;
48         memset(o,0,sizeof(o));
49         scanf("%s",str);
50         len = strlen(str);
51         flag = 0;
52         for(i = 0; i < len; i ++)
53         {
54             for(j = i+1; j < len; j ++)
55             {
56                 if(str[i] == str[j])
57                     for(k = j+1; k < len; k ++)
58                     {
59                         if(str[j] == str[k])
60                         {
61                             if(judge(i,j,k))
62                             {
63                                 flag = 1;
64                                 o[str[i]+0] ++;
65                             }
66                         }
67                     }
68             }
69         }
70         if(!flag) 
71         {
72             printf("LOOOOOOOOSER!
");
73             continue;
74         }
75         for(i = 0;i <= 128;i ++)
76         {
77             if(o[i])
78             {
79                 for(j = 1;j <= o[i];j ++)
80                 printf("%c",i);
81             }
82         }
83         printf("
");
84     }
85     return 0;
86 }