假设确定了最终所得向量的方向，则应该选择所有在该方向上投影为正的向量。按极角序排序后这显然是一段连续区间。最终向量方向很难枚举，但对于某个向量，在其上投影为正的向量与其夹角范围是(-π/2,π/2)，所以只要枚举所有极角差不超过π的极长区间就可以了。这里的区间不是向量区间而是极角区间，相当于一条过原点的直线在旋转，所以双指针移动时每个向量区间都要更新答案。为了方便倍长向量数组，注意这样对于增加的那一半，极角需要加上2π。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
int x=,f=;char c=getchar();
while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
return x*f;
}
int n;
const double PI=3.14159266;
ll ans;
struct vector
{
int x,y;double angle;
bool operator <(const vector&a) const
{
return angle<a.angle;
}
vector operator +(const vector&a) const
{
return (vector){x+a.x,y+a.y};
}
vector operator -(const vector&a) const
{
return (vector){x-a.x,y-a.y};
}
ll len(){return 1ll*x*x+1ll*y*y;}
}a[N<<];
int main()
{
#ifndef ONLINE_JUDGE
freopen("bzoj5099.in","r",stdin);
freopen("bzoj5099.out","w",stdout);
const char LL[]="%I64d\n";
#else
const char LL[]="%lld\n";
#endif
n=read();
for (int i=;i<=n;i++) a[i].x=read(),a[i].y=read(),a[i].angle=atan2(a[i].x,a[i].y);
sort(a+,a+n+);
for (int i=n+;i<=n*;i++) a[i]=a[i-n],a[i].angle+=*PI;
vector cur=(vector){,};
int x=;
for (int i=;i<=n;i++)
{
while (x<n*&&a[x+].angle-a[i].angle<=PI) ans=max(ans,(cur=cur+a[++x]).len());
ans=max(ans,(cur=cur-a[i]).len());
}
cout<<ans;
return ;
}