大致题意：

　　给出一个多边形，问你有多少种放法可以使得多边形稳定得立在平面上。

 

　　先对多边形求重心，在求凸包，枚举凸包的边，如果重心没有在边的范围内，则不行

　　判断是否在范围内可用点积来判断

　　　 
  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<queue>
  6 #include<set>
  7 #include<map>
  8 #include<stack>
  9 #include<time.h>
 10 #include<cstdlib>
 11 #include<cmath>
 12 #include<list>
 13 using namespace std;
 14 #define MAXN 100100
 15 #define eps 1e-9
 16 #define For(i,a,b) for(int i=a;i<=b;i++) 
 17 #define Fore(i,a,b) for(int i=a;i>=b;i--) 
 18 #define lson l,mid,rt<<1
 19 #define rson mid+1,r,rt<<1|1
 20 #define mkp make_pair
 21 #define pb push_back
 22 #define cr clear()
 23 #define sz size()
 24 #define met(a,b) memset(a,b,sizeof(a))
 25 #define iossy ios::sync_with_stdio(false)
 26 #define fre freopen
 27 #define pi acos(-1.0)
 28 #define inf 1e6+7
 29 #define Vector Point
 30 const int Mod=1e9+7;
 31 typedef unsigned long long ull;
 32 typedef long long ll;
 33 int dcmp(double x){
 34     if(fabs(x)<=eps) return 0;
 35     return x<0?-1:1;
 36 }
 37 struct Point{
 38     double x,y;
 39     Point(double x=0,double y=0):x(x),y(y) {}
 40     bool operator < (const Point &a)const{
 41         if(x==a.x) return y<a.y;
 42         return x<a.x;
 43     }
 44     Point operator - (const Point &a)const{
 45         return Point(x-a.x,y-a.y);
 46     }
 47     Point operator + (const Point &a)const{
 48         return Point(x+a.x,y+a.y);
 49     }
 50     Point operator * (const double &a)const{
 51         return Point(x*a,y*a);
 52     }
 53     Point operator / (const double &a)const{
 54         return Point(x/a,y/a);
 55     }
 56     void read(){
 57         scanf("%lf%lf",&x,&y);
 58     }
 59     void out(){
 60         cout<<"debug: "<<x<<" "<<y<<endl;
 61     }
 62     bool operator == (const Point &a)const{
 63         return dcmp(x-a.x)==0 && dcmp(y-a.y)==0;
 64     }
 65 };
 66 double Dot(Vector a,Vector b) {
 67     return a.x*b.x+a.y*b.y;
 68 }
 69 double dis(Vector a) {
 70     return sqrt(Dot(a,a));
 71 }
 72 double Cross(Point a,Point b){
 73     return a.x*b.y-a.y*b.x;
 74 }
 75 int ConvexHull(Point *p,int n,Point *ch){
 76     int m=0;
 77     For(i,0,n-1) {
 78         while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
 79         ch[m++]=p[i];
 80     }
 81     int k=m;
 82     Fore(i,n-2,0){
 83         while(m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
 84         ch[m++]=p[i];
 85     }
 86     if(n>1) m--;
 87     return m;
 88 }
 89 int n,m;
 90 Point p[100005];
 91 Point ch[100005];
 92 Point cp;
 93 void solve(){
 94     scanf("%d",&n);
 95     For(i,0,n-1) p[i].read();
 96     double tar=0,ar;
 97     cp.x=0;cp.y=0;
 98     For(i,2,n-1){
 99         ar=Cross(p[i]-p[0],p[i-1]-p[0])/2;
100         tar+=ar;
101         cp=cp+(p[i]+p[i-1]+p[0])*ar;
102     }
103     cp=cp/tar/3;
104     sort(p,p+n);
105     m=ConvexHull(p,n,ch);
106     int ans=0;
107     For(i,0,m-1) {
108         int nxt=(i+1)%m;
109         if(dcmp(Dot(ch[i]-ch[nxt],cp-ch[nxt]))>0 && dcmp(Dot(ch[nxt]-ch[i],cp-ch[i]))>0) ans++;
110     }
111     printf("%d\n",ans);
112 }
113 int main(){
114 //    fre("in.txt","r",stdin);
115     int t=0;
116     cin>>t;
117     For(i,1,t) solve();
118     return 0;
119 }
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