POJ 1244 Slots of Fun(计算几何)

题目链接

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!\n");

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("\n");

84     }

85     return 0;

86 }