[Gym

大致题意：

　　给出一个凸包，以及凸包内的两个点p1,p2，求有多少条经过凸包顶点的直线能够将凸包分割为两部分，且给出的两点分别属于不同的部分

　　

　　枚举凸包的顶点，二分求出p1,p2线段左边的最大坐标L以及右边的最小坐标R，则答案为R-L-1的累加和除以2

　　注意文件输入，输出

　　　

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<cstdlib>
#include<cmath>
#include<list>
using namespace std;
#define MAXN 100100
#define eps 1e-9
#define For(i,a,b) for(int i=a;i<=b;i++) 
#define Fore(i,a,b) for(int i=a;i>=b;i--) 
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mkp make_pair
#define pb push_back
#define cr clear()
#define sz size()
#define met(a,b) memset(a,b,sizeof(a))
#define iossy ios::sync_with_stdio(false)
#define fre freopen
#define pi acos(-1.0)
#define inf 1e6+7
#define Vector Point
const int Mod=1e9+7;
typedef unsigned long long ull;
typedef long long ll;
int dcmp(double x){
    if(fabs(x)<=eps) return 0;
    return x<0?-1:1;
}
struct Point{
    double x,y;
    int id;
    Point(double x=0,double y=0):x(x),y(y) {}
    bool operator < (const Point &a)const{
        if(x==a.x) return y<a.y;
        return x<a.x;
    }
    Point operator - (const Point &a)const{
        return Point(x-a.x,y-a.y);
    }
    Point operator + (const Point &a)const{
        return Point(x+a.x,y+a.y);
    }
    Point operator * (const double &a)const{
        return Point(x*a,y*a);
    }
    Point operator / (const double &a)const{
        return Point(x/a,y/a);
    }
    void read(){
        scanf("%lf%lf",&x,&y);
    }
    void out(){
        cout<<"debug: "<<x<<" "<<y<<endl;
    }
    bool operator == (const Point &a)const{
        return dcmp(x-a.x)==0 && dcmp(y-a.y)==0;
    }
};
double Dot(Vector a,Vector b) {
    return a.x*b.x+a.y*b.y;
}
double dis(Vector a) {
    return sqrt(Dot(a,a));
}
double Cross(Point a,Point b){
    return a.x*b.y-a.y*b.x;
}
int n;
Point p[200005];
int ls,rs;
int chk1(Point p1,Point p2,int pos){
    int l=pos+1,r=pos+n-1;
    while(l<r){
        int mid=l+r>>1;
        if(dcmp(Cross(p[mid]-p[pos],p1-p[pos]))<=0 && dcmp(Cross(p[mid]-p[pos],p2-p[pos])<=0)) r=mid;
        else l=mid+1;
    }
    return l;
}
int chk2(Point p1,Point p2,int pos){
    int l=pos+1,r=pos-1+n;
    while(l<r){
        int mid=l+r+1>>1;
        if(dcmp(Cross(p[mid]-p[pos],p1-p[pos]))>=0 && dcmp(Cross(p[mid]-p[pos],p2-p[pos]))>=0) l=mid;
        else r=mid-1;
    }
    return l;
}
Point p1,p2;
void solve(){
    For(i,1,n) p[i].read();
    p1.read();p2.read();
    ll ans=0;
    For(i,1,n) p[i+n]=p[i];
    For(i,1,n){
        ls=chk1(p1,p2,i);
        rs=chk2(p1,p2,i);
        if(ls!=rs)ans+=abs(ls-rs)-1;
    }
    cout<<ans/2<<endl;
}
int main(){
//    fre("in.txt","r",stdin);
    freopen("kingdom.in","r",stdin);
    freopen("kingdom.out","w",stdout);
    int t=0;
     while(~scanf("%d",&n) &&n)    solve();
    return 0;
}