Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are forced to save money on buying fence posts by using trees as fence posts wherever possible. Given the locations of some trees, you are to help farmers try to create the largest pasture that is possible. Not all the trees will need to be used.

However, because you will oversee the construction of the pasture yourself, all the farmers want to know is how many cows they can put in the pasture. It is well known that a cow needs at least 50 square metres of pasture to survive.

The first line of input contains a single integer, n (1 ≤ n ≤ 10000), containing the number of trees that grow on the available land. The next n lines contain the integer coordinates of each tree given as two integers x and yseparated by one space (where -1000 ≤ x, y ≤ 1000). The integer coordinates correlate exactly to distance in metres (e.g., the distance between coordinate (10; 11) and (11; 11) is one metre).

You are to output a single integer value, the number of cows that can survive on the largest field you can construct using the available trees.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct pnt{    int x,y;    pnt(){x=y=0;}    pnt(int tx,int ty){x=tx;y=ty;}    pnt operator -(pnt p2){return pnt(x-p2.x,y-p2.y);}    pnt operator +(pnt p2){return pnt(x+p2.x,y+p2.y);}    bool operator <(pnt p2)const {if(x!=p2.x)return x<p2.x;return y<p2.y;}    bool operator >(pnt p2)const {if(x!=p2.x)return x>p2.x;return y>p2.y;}    bool operator ==(pnt p2)const {return x==p2.x&&y==p2.y;}    int dot(pnt p2){return x*p2.x+y*p2.y;}//点积    int det(pnt p2){return x*p2.y-y*p2.x;}//叉积};const int maxn=1e4+2;pnt p[maxn],h[maxn];int n,m;void convexHull(){    sort(p,p+n);    m=0;    for(int i=0;i<n;i++){//计算上凸包        while(m>1&&(h[m-1]-h[m-2]).det(p[i]-h[m-2])<=0){m--;}        h[m++]=p[i];    }    int tm=m;    for(int i=n-2;i>=0;i--){//计算下凸包        while(m>tm&&(h[m-1]-h[m-2]).det(p[i]-h[m-2])<=0){m--;}        h[m++]=p[i];    }    if(n>1)m--;}double calArea(){    double ans=0;    for(int i=0;i<m;i++){        ans+=(double)((h[i]-h[0]).det(h[(i+1)%m]-h[0]))/2;//使用叉积计算面积    }    return ans;}int main(){    scanf("%d",&n);    for(int i=0;i<n;i++)scanf("%d%d",&p[i].x,&p[i].y);    convexHull();    int ans=(int)(calArea()/50.0);    printf("%d\n",ans);    return 0;}