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 ;

 }