题意：判断点在多边形内外，用来试模版的

#include <iostream>
#include <stdio.h>
#include <math.h>

typedef double DIY;

const DIY EPS = 0;
const DIY N = 1000005;

using namespace std;
struct Point
{
    DIY x,y;
};
struct Segment
{
    Point a;
    Point b;
};
typedef Point point;
Point p[100];
DIY cross(Point A,Point B,Point C)
{
    return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);
}
bool IsOnline(Point p,Segment s)
{
    if(fabs(cross(s.a,p,s.b))<=EPS&&min(s.a.x,s.b.x)<=p.x&&
            p.x<=max(s.a.x,s.b.x)&&min(s.a.y,s.b.y)<=p.y&&p.y<=max(s.a.y,s.b.y))
        return true;
    return false;
}
bool Segment_crossing(Segment u,Segment v)  /*判断两线段是否相交*/
{
    return((max(u.a.x,u.b.x)>=min(v.a.x,v.b.x))&&
           (max(v.a.x,v.b.x)>=min(u.a.x,u.b.x))&&
           (max(u.a.y,u.b.y)>=min(v.a.y,v.b.y))&&
           (max(v.a.y,v.b.y)>=min(u.a.y,u.b.y))&&
           (cross(v.a,u.b,u.a)*cross(u.b,v.b,u.a)>=0)&&
           (cross(u.a,v.b,v.a)*cross(v.b,u.b,v.a)>=0));
}
bool Inside(Point t,Point p[],int n)     /*判断点p是否在以边点顺序排列的点集s所围成的多边形内*/
{
    Segment seg;
    Segment pp;
    int num=0;
    Point MM;          /*定义无穷点*/
    MM.y=t.y;
    MM.x=N;
    pp.a=t;
    pp.b=MM;           /*构造一条平行于x轴的以p为端点的右射线pp*/
    for(int i=0; i<n; i++)
    {
        seg.a=p[i];
        seg.b=p[(i+1)%n];
        if(IsOnline(t,seg))
            return 1;
        if(fabs(seg.a.y-seg.b.y)<EPS)
            continue;
        if(IsOnline(seg.a,pp))
        {
            if(seg.a.y>seg.b.y)
                num++;
        }
        else if(IsOnline(seg.b,pp))
        {
            if(seg.b.y>seg.a.y)
                num++;
        }
        else if(Segment_crossing(pp,seg))
            num++;
    }
    if(num%2==1)
        return true;
    return false;
}
int main()
{
    Point t;
    int n,m;
    while(cin>>n)
    {
        for(int i=0; i<n; i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        cin>>m;
        while(m--)
        {
            scanf("%lf%lf",&t.x,&t.y);
            if(Inside(t,p,n)) puts("Yes");
            else              puts("No");
        }
    }
    return 0;
}