HDU4195 Regular Convex Polygon (正多边形、外接圆)

题意：

给你正n边形上的三个点，问n最少为多少

思路：

三个点在多边形上，所以三个点的外接圆就是这个正多边形的外接圆，余弦定理求出每个角的弧度值，即该角所对边的圆周角，该边对应的圆心角为圆心角的二倍。同时这个圆心角应为正多边形的每条边对应圆心角的整数倍，即2*pi/i的整数倍，遍历for i 1 to 1000 ，判断A*i/pi是否均为整数即可

坑点：

注意判断double是否为整数：return a-(int)(a+eps);

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 5e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

struct point{
    double x, y;
    point(double x = , double y = ):x(x), y(y){}
};
double length(point a, point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool ok(double a){
    return a >  &&fabs(a-(int)(a+eps)) < eps;
}
int main(){
    point a, b, c;
    while(scanf("%lf %lf", &a.x, &a.y) != ){
        scanf("%lf %lf %lf %lf", &b.x, &b.y, &c.x, &c.y);
        double ab = length(a, b);
        double ac = length(a, c);
        double bc = length(b, c);
        double A = acos((ab*ab+ac*ac-bc*bc)/(2.0*ab*ac));
        double B = acos((ab*ab+bc*bc-ac*ac)/(2.0*ab*bc));
        double C = acos((ac*ac+bc*bc-ab*ab)/(2.0*ac*bc));
        int flg = ;
        for(int i = ; i <= ; i++){
            if(ok(A*i/pi) && ok(B*i/pi) && ok(C*i/pi)){
                flg = i;
                break;
            }
        }
        printf("%d\n", flg);

    }
    return ;
}