题意：
给出平面上n个点，问能在其中选出6个点，组成两个三角形，使得其互不相交
问有多少种选法
大致思路
考虑枚举一条直线，将所有得点分为左右两部分，其中有两个点在直线上，
以这两个点为顶点，分别统计左边有多少个三角形，右边有多少个三角形即可。
枚举直线时，枚举一个点，然后对其他点极角排序，
枚举另一个点，可快速算出该直线左右两边的点的数量(二分，叉积），组合一下即可。
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*/ #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 10000006
#define eps 0
#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 fr freopen
#define pi acos(-1.0)
#define Vector Point
#define fir first
#define sec second
#define it_s_too_hard main
const long long inf=1LL<<;
const int iinf=<<;
const double dinf=1e17;
const int Mod=1e9+;
typedef long long ll;
typedef long double ld;
struct Point{
ll x,y;
double ang;
Point(ll x=,ll y=):x(x),y(y) {}
Point operator - (const Point &a)const { return Point(x-a.x,y-a.y);}
};
ll Cross(Vector a,Vector b){ return a.x*b.y-a.y*b.x;}
int n;
Point o;
Vector v1,v2;
bool cmp(Point a,Point b){
return a.ang<b.ang;
}
Point p[],tp[];
void I_can_t_solve_it(){
cin>>n;
ll ans=,cnt1,l,r,ct;
For(i,,n) cin>>p[i].x>>p[i].y;
For(i,,n){
For(j,,n) {
if(j==i) continue;
if(j>i) tp[j-]=p[j]-p[i],tp[j-].ang=atan2(tp[j-].y,tp[j-].x);
else tp[j]=p[j]-p[i],tp[j].ang=atan2(tp[j].y,tp[j].x);
}
ct=n-;cnt1=;
sort(tp+,tp+ct+,cmp);
For(j,,ct) tp[j+ct]=tp[j];
For(j,,ct){
if(j>cnt1) cnt1=j;
if(cnt1>ct*) break;
while(cnt1-j+<=ct && cnt1<ct* && Cross(tp[j],tp[cnt1+])>) cnt1++;
l=cnt1-j;r=n--l-;
ans+=l*(l-)*r*(r-)/;
}
}
cout<<ans/<<endl; }
int it_s_too_hard(){
int t=;
iossy;
For(i,,t){
I_can_t_solve_it();
}
return ;
}