/*
poj3348 Cows 凸包+多边形面积 水题
floor向下取整，返回的是double
*/
#include
#include
#include  
using namespace std;
const double eps = 1e-8;  
struct point
{
	double x,y;
};
int n;
point dian[10000+10],zhan[10000+10];
//////////////////////////////////////////////////
point *mo_dian;
double mo_distance(point p1,point p2)
{
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
double mo_xmult(point p2,point p0,point p1)//p1在p2左返回负，在右边返回正
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

bool mo_ee(double x,double y)
{
	double ret=x-y;
	if(ret<0) ret=-ret;
	if(ret y + eps;} // x > y   
bool mo_ll(double x,double y)  {   return x < y - eps;} // x < y   
bool mo_ge(double x,double y) {   return x > y - eps;} // x >= y   
bool mo_le(double x,double y) {   return x < y + eps;}     // x <= y   

bool mo_cmp(point a,point b)   // 第一次排序   
{  
    if( mo_ee(a.y ,b.y ) )  
        return mo_ll(a.x, b.x);  
    return mo_ll(a.y,b.y);  
}  
bool mo_cmp1(point a,point b)  // 第二次排序   
{  
    double len = mo_xmult(b,mo_dian[0],a);  
    if( mo_ee(len,0.0) )  
        return mo_ll(mo_distance(mo_dian[0],a),mo_distance(mo_dian[0],b));  
    return mo_gg(len,0.0);  
}  
int mo_graham(int n,point *dian,point *stk)
{
	int i,top=1;
	mo_dian=dian;
	sort(mo_dian,mo_dian+n,mo_cmp);  
    sort(mo_dian+1,mo_dian+n,mo_cmp1);  
	stk[0]=mo_dian[0];  
    stk[1]=mo_dian[1];  
	for(i=2;i0&&mo_xmult(mo_dian[i],stk[top-1],stk[top])<=0) --top;  
        stk[++top]=mo_dian[i];  
    }  
    return top+1;  
}
double mo_area_polygon(point *dian,int n)
{
	int i;
	point yuan;
	yuan.x=yuan.y=0;
	double ret=0;
	for(i=0;i