首先求多邊形面積，這個比較簡單，用的就是把一個多邊形劃分為多個三角形，然後求三角形面積。

程式碼：

double Cross(Vector A,Vector B) { return (A.x*B.y-A.y*B.x); }
double ConvexPolygonArea(Point* p,int n)//多邊形面積,，點按順序
{
    double area=0;
    for(int i=1;i<n-1;i++)
        area+=Cross(p[i]-p[0],p[i+1]-p[0]);
    return area/2;
}

求凸包的演算法有很多，這裡給出Andrew演算法。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
using namespace std;
#define Del(a,b) memset(a,b,sizeof(a))
const int N = 1010;
const double esp = 1e-10;
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y) {}
};
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 * (Vector A,double p) { 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);
}
int dcmp(double x)  //
{
    if(fabs(x)<esp) 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;
}
///計算點積，及向量長度，及向量夾角
double Dot(Vector A,Vector B) { return A.x*B.x+A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A,A)); }
double Angle(Vector A,Vector B) { return acos(Dot(A,B))/Length(A)/Length(B); }
//計算叉積，向量逆時針旋轉
double Cross(Vector A,Vector B) { return (A.x*B.y-A.y*B.x); }
double Area2(Vector A,Vector B,Vector C)  { return Cross(B-A,C-A); }
Vector Rotate(Vector A,double rad)
{
    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
int cmp(Point a,Point b)
{
    if(a.x!=b.x)
        return a.x<b.x;
    if(a.y!=b.y)
        return a.y<b.y;
}
double ConvexPolygonArea(Point* p,int n)//多邊形面積,，點按順序
{
    double area=0;
    for(int i=1;i<n-1;i++)
        area+=Cross(p[i]-p[0],p[i+1]-p[0]);
    return area/2;
}
int ConvexHull(Point *p,Point *ch,int n)//求凸包
{
    sort(p,p+n);
    int i,m=0,k;
    for(i=0;i<n;i++)
    {
        while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
        ch[m++]=p[i];
    }
    k=m;
    for(i=n-2;i>=0;i--)
    {
        while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
        ch[m++]=p[i];
    }
    if(n>1)m--;
    return m;
}
int main()
{
    int n,k;
    Point a[N],ch[N];
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
            scanf("%lf%lf",&a[i].x,&a[i].y);
        int num=ConvexHull(a,ch,n);
        double ans=ConvexPolygonArea(ch,num);
        ans/=50.0;
        printf("%d\n",(int)ans);
    }
    return 0;
}