题目
本身是一道水题，但我调了好久，最后发现 y 1 / x 1 &lt; y 2 / x 2 y1/x1&lt;y2/x2 y1/x1<y2/x2与 y 1 ∗ x 2 &lt; y 2 ∗ x 1 y1*x2&lt;y2*x1 y1∗x2<y2∗x1并不等价

#include<bits/stdc++.h>
using namespace std;
struct node{
	int x,y;
}a[2001],b[2001];
long long ans;
int n,i,s,now,k;
inline char gc(){
	static char buf[100000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int rd(){
	int x=0,fl=1;char ch=gc();
	for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
	for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
	return x*fl;
}
inline void wri(int a){if(a<0)a=-a,putchar('-');if(a>=10)wri(a/10);putchar(a%10|48);}
inline void wln(int a){wri(a);puts("");}
bool cmp(node a,node b){
	return (a.x==0?1e99:1.0*a.y/a.x)<(b.x==0?1e99:1.0*b.y/b.x);
}
/*无精度误差 bool cmp(node a,node b){ if (a.x==0) return 0; if (b.x==0) return 1; return (a.y*b.x<b.y*a.x)^(a.x*b.x<0); } */
int main(){
	n=rd();
	for (i=0;i<n;i++) a[i].x=rd(),a[i].y=rd(),b[i]=a[i];
	for (now=0;now<n;now++){
		k=0;
		for (i=now+1;i<n;i++) a[k].x=b[i].x-b[now].x,a[k++].y=b[i].y-b[now].y;
		sort(a,a+k,cmp);
		s=1;
		for (i=1;i<k;i++)
			if (a[i].y*a[i-1].x==a[i-1].y*a[i].x) s++;
			else ans+=s*(s-1)/2,s=1;
		ans+=s*(s-1)/2;
	}
	printf("%lld",1ll*n*(n-1)*(n-2)/6-ans);
}