我们把所有的直线按斜率从小到大排序，然后用单调栈维护
发现，如果当前直线与(st[top-1])直线的交点的横坐标大于等于与(st[top])的交点的横坐标，当前直线可以覆盖掉(st[top])
这个可以画图理解一下
//minamoto
#include<bits/stdc++.h>
#define rint register int
using namespace std;
#define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
int read(){
int res,f=1;char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
const int N=50005;
struct node{
int a,b,id;
inline bool operator <(const node &q)const{return a==q.a?b>q.b:a<q.a;}
}p[N];int n,st[N],top;
inline double solve(int i,int j){return (double)(p[i].b-p[j].b)/(p[j].a-p[i].a);}
int main(){
// freopen("testdata.in","r",stdin);
n=read();for(rint i=1;i<=n;++i)p[i].a=read(),p[i].b=read(),p[i].id=i;
sort(p+1,p+1+n),st[top=0]=1;
for(rint i=2;i<=n;++i){
if(p[i].a==p[i-1].a)continue;
while(top&&solve(st[top],i)<=solve(st[top-1],i))--top;
st[++top]=i;
}for(rint i=0;i<=top;++i)st[i]=p[st[i]].id;
sort(st,st+1+top);for(rint i=0;i<=top;++i)printf("%d ",st[i]);return 0;
}