#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define INF 0x3f3f3f3f3f3f3f3f
#define MOD 1000000007
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
const long double pi = acos(-1.0);

pair<ll,ll>xy[3010];
pair<ll,ll>xysort[3010];
int n;

template<class T1, class T2>
inline const pair<T1, T2> operator-(const pair<T1, T2>&p1, const pair<T1, T2>&p2) {
  return {
    p1.first - p2.first,
      p1.second - p2.second
  };
}

ll cross(pair<ll,ll> vec1,pair<ll,ll> vec2){
  auto [x1,y1] = vec1;
  auto [x2,y2] = vec2;
  return x1*y2-x2*y1;
}

int main() {
  ll n;
  scanf("%lld",&n);
  rep(i,0,n){
    ll x,y;
    scanf("%lld %lld",&x,&y);
    xy[i] = {x,y};
  }
  ll res = n * ((n-1) * (n-2) * (n-3) * (n-4) / (1*2*3*4));
  rep(i,0,n){
    rep(j,0,n){
      if(i == j)continue;
      xysort[j>i?j-1:j] = xy[j];
    }
    sort(xysort+1,xysort+n-1,
        [i](const pair<ll,ll>p0,const pair<ll,ll>p1) -> bool{
        ll cross0 = cross(p0-xy[i],xysort[0]-xy[i]);
        ll cross1 = cross(p1-xy[i],xysort[0]-xy[i]);
        if(cross0 >=0 && cross1 < 0){
          return true;
        }else if(cross0 < 0 && cross1 >= 0){
          return false;
        }
        ll cross01=cross(p0-xy[i],p1-xy[i]);
        return cross01 <= 0;
        });
    int index = 1;
    // 四个点的起始点,首尾游标
    rep(j,0,n-1){
      while(index < j + n-1){
        ll crossij = cross(xysort[index%(n-1)]-xy[i],xysort[j]-xy[i]);
        if(crossij >= 0 ){
          index++;
        }else{
          break;
        }
      }
      // C_cnt^3
      ll cnt = index - j - 1;
      res -= (cnt * (cnt - 1) * (cnt - 2)) / (1*2*3);
    }
  }
  printf("%lld\n",res);
  return 0;
}

总结

1是想到反向计算，我一直在想算三角形然后带其它点，再容斥，就没想出来

2是浮点数精度，哎自闭，还是整数香