【SCOI2019】—DAY2T1 湖之精灵的游戏（凸包+二分）

考场$debuff$太严重了
下来一眼沙比提

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考虑其实就是
求一段区间
$Max|\sum_{i\in [l,r]}(x_i,y_i)*(x,y)|$

因为叉积不同方向会有正负
由于有绝对值，可以拆成最大前缀和减去最小前缀和
先只考虑最大前缀和
即$Max \sum{x_i}*y-\sum{y_i}*x$
考虑2个前缀$i,j$，$i$比$j$大，则
$(\sum{x_i}-\sum{x_j})*y-(\sum{y_i}-\sum{y_j})*x>0$
化一下就变成了
$\frac{(\sum{y_i}-\sum{y_k})}{(\sum{x_i}-\sum{x_j})}<\frac y x$

就是一个很裸地斜率式子了，每次二分即可
复杂度$O(nlogn)$

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ob==ib)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
const int N=1000005;
int X[N],Y[N],n,m1,m2,q;
struct point{
	int x,y;
	point(int _x=0,int _y=0):x(_x),y(_y){}
	friend inline point operator +(const point &a,const point &b){
		return point(a.x+b.x,a.y+b.y);
	}
	friend inline point operator -(const point &a,const point &b){
		return point(a.x-b.x,a.y-b.y);
	}
	friend inline int operator *(const point &a,const point &b){
		return (a.x*b.y-a.y*b.x);
	}
}p[N],up[N],down[N];
inline void graham(){
	up[1]=down[1]=point(0,0),up[2]=down[2]=p[1],m1=m2=2;
	for(int i=2;i<=n;i++){
		while(m1>=2&&(up[m1]-up[m1-1])*(p[i]-up[m1-1])>=0)m1--;
		up[++m1]=p[i];
		while(m2>=2&&(p[i]-down[m2-1])*(down[m2]-down[m2-1])>=0)m2--;
		down[++m2]=p[i];
	}
}
inline double calc(const point &a,double k){
	return a.y-a.x*k;
}
inline double checkup(double k){
	int l=1,res=l,r=m1;
	while(l<=r){
		int mid=(l+r)>>1;
		if(calc(up[mid],k)>=calc(up[mid+1],k))r=mid-1,res=mid;
		else l=mid+1;
	}
	return calc(up[res],k);
}
inline double checkdown(double k){
	int l=1,res=l,r=m2;
	while(l<=r){
		int mid=(l+r)>>1;
		if(calc(down[mid],k)<=calc(down[mid+1],k))r=mid-1,res=mid;
		else l=mid+1;
	}
	return calc(down[res],k);
}
signed main(){
	n=read();
	for(int i=1;i<=n;i++)X[i]=read(),Y[i]=read();
	for(int i=1;i<=n;i++)p[i]=p[i-1]+point(X[i],Y[i]);
	graham();
	q=read();
	for(int i=1;i<=q;i++){
		int x=read(),y=read();
		double k=1.0*y/x;
		double f1=checkup(k);
		double f2=checkdown(k);
		cout<<(int)round(x*(f1-f2))<<'\n';
	}
}