HDU 4195 Regular Convex Polygon

思路：三角形的圆心角可以整除(2*pi)/n

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<queue>
 #include<stack>
 #include<algorithm>
 using namespace std;
 #define clc(a,b) memset(a,b,sizeof(a))
 #define inf 0x3f3f3f3f
 const int N=;
 #define LL long long
 const double eps = 1e-;
 const double pi = acos(-);
 // inline int r(){
 //     int x=0,f=1;char ch=getchar();
 //     while(ch>'9'||ch<'0'){if(ch=='-') f=-1;ch=getchar();}
 //     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 //     return x*f;
 // }
 struct  Point{
     double x,y;
     Point(double a=,double b=):x(a),y(b){}
 };

 double angle(Point a,Point b,Point c){
     b.x-=a.x,b.y-=a.y;
     c.x-=a.x,c.y-=a.y;
     return acos((b.x*c.x+b.y*c.y)/(hypot(b.x,b.y)*hypot(c.x,c.y)));
 }

 bool dcmp(double f,int n){
     return fabs(f * n - round(f * n)) < eps;
 }

 int main(){
     while(){
         Point p[];
         for(int i=;i<;i++)
             if(scanf("%lf%lf",&p[i].x,&p[i].y)!=) return ;
         double a=angle(p[],p[],p[])/pi;
         double b=angle(p[],p[],p[])/pi;
         for(int i=;i<=;i++){
             if(dcmp(a,i)&&dcmp(b,i)){
                 printf("%d\n",i);
                 break;
             }
         }
     }
     return ;
 }