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            using namespace std;const double eps = 1e-8;const double PI = acos(-1.0);const double INF = 1000000000000000.000;struct Point{ double x,y; Point(double x=0, double y=0) : x(x),y(y){ } //构造函数};typedef Point Vector;Vector operator + (Vector A , Vector B){ return Vector(A.x+B.x,A.y+B.y);}Vector operator - (Vector A , Vector B){ return Vector(A.x-B.x,A.y-B.y);}Vector operator * (double p,Vector A){ return Vector(A.x*p,A.y*p);}Vector operator / (Vector A , double p){ return Vector(A.x/p,A.y/p);}bool operator < (const Point& a,const Point& b){ return a.x < b.x ||( a.x == b.x && a.y < b.y);}inline int dcmp(double x){ if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;}bool operator == (const Point& a, const Point& b){ return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;}///向量（x,y）的极角用atan2(y,x);inline double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; }inline double Length(Vector A) { return sqrt(Dot(A,A)); }inline double Angle(Vector A, Vector B) { return acos(Dot(A,B) / Length(A) / Length(B)); }inline double Cross(Vector A, Vector B) { return A.x*B.y - A.y * B.x; }//凸包：/**Andrew算法思路：首先按照先x后y从小到大排序（这个地方没有采用极角逆序排序，所以要进行两次扫描），删除重复的点后得到的序列p1,p2.....,然后把p1和p2放到凸包中。从p3开始，当新的点在凸包“前进”方向的左边时继续，否则依次删除最近加入凸包的点，直到新点在左边；**///Goal[]数组模拟栈的使用；int ConvexHull(Point* P,int n,Point* Goal){ sort(P,P+n); int m = unique(P,P+n) - P; //对点进行去重； int cnt = 0; for(int i=0;i
            
             1 && dcmp(Cross(Goal[cnt-1]-Goal[cnt-2],P[i]-Goal[cnt-2])) <= 0) cnt--; //如果希望在凸包的边上有输入点，把<= 换成 <. Goal[cnt++] = P[i]; } int temp = cnt; for(int i=m-2;i>=0;i--){ //逆序求上凸包； while(cnt>temp && dcmp(Cross(Goal[cnt-1]-Goal[cnt-2],P[i]-Goal[cnt-2])) <= 0) cnt--; Goal[cnt++] = P[i]; } if(cnt > 1) cnt--; //减一为了去掉首尾重复的； return cnt;}bool IsPointOnSegment(Point P,Point A,Point B){ return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(P-A,P-B)) < 0;}/*************************************分 割 线*****************************************/const int maxn = 1005;Point P[maxn];Point Goal[maxn];int main(){ freopen("E:\\acm\\input.txt","r",stdin); int T; cin>>T; while(T--){ int n; cin>>n; for(int i=0;i
             
               n || n < 6){ printf("NO\n"); continue; } cout<
              
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