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=10010;
#define LL long long
const double eps = 1e-5;
const double pi = acos(-1);
// 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=0,double b=0):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(1){
        Point p[3];
        for(int i=0;i<3;i++)
            if(scanf("%lf%lf",&p[i].x,&p[i].y)!=2) return 0;
        double a=angle(p[0],p[1],p[2])/pi;
        double b=angle(p[1],p[2],p[0])/pi;
        for(int i=3;i<=1000;i++){
            if(dcmp(a,i)&&dcmp(b,i)){
                printf("%d
",i);
                break;
            }
        }
    }
    return 0;
}