CF1041F Ray in the tube

原题链接：http://codeforces.com/contest/1041/problem/F

Ray in the tube

You are given a tube which is reflective inside represented as two non-coinciding, but parallel to $Ox$ lines. Each line has some special integer points — positions of sensors on sides of the tube.

You are going to emit a laser ray in the tube. To do so, you have to choose two integer points $A$ and $B$ on the first and the second line respectively (coordinates can be negative): the point $A$ is responsible for the position of the laser, and the point $B$ — for the direction of the laser ray. The laser ray is a ray starting at $A$ and directed at $B$ which will reflect from the sides of the tube (it doesn’t matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.

Examples of laser rays. Note that image contains two examples. The $3$ sensors (denoted by black bold points on the tube sides) will register the blue ray but only $2$ will register the red.

Calculate the maximum number of sensors which can register your ray if you choose points $A$ and $B$ on the first and the second lines respectively.

Input

The first line contains two integers $n$ and $y_1 (1≤n≤10^5, 0≤y_1≤10^9)$ — number of sensors on the first line and its $y$ coordinate.

The second line contains $n$ integers $a_1,a_2,\cdots,a_n (0≤a_i≤10^9)$ — $x$ coordinates of the sensors on the first line in the ascending order.

The third line contains two integers $m$ and $y_2 (1≤m≤10^5, y_1<y_2≤10^9)$ — number of sensors on the second line and its $y$ coordinate.

The fourth line contains $m$ integers $b_1,b_2,\cdots,b_m (0≤b_i≤10^9)$ — $x$ coordinates of the sensors on the second line in the ascending order.

Output

Print the only integer — the maximum number of sensors which can register the ray.

Example

input

3 1
1 5 6
1 3
3

output

Note

One of the solutions illustrated on the image by pair $A_2$ and $B_2$ .

题解

如果在实数域做的话，我们可以从一个无限接近于 $90^\circ$ 的角度发出射线，这样整个管子都是射线岂不是很棒？

然而放到整数域的话，我们能做到的最极限角度就是步长为 $1$ 了，这样由于反射问题，在同一侧管壁上射线就是两个点两个点的跳的，导致有一半的整点达不到：

谁能填这些空呢？我们试试步长为 $2$ 的射线：

发现只填了一半的空，再试试步长为 $3$ 的，发现被步长为 $1$ 完虐，而步长为 $4$ 的射线又可以填掉一半的空。。。以此类推，可以发现只有步长为 $2^n$ 的射线可以到达一些其他线到不了的点，除此之外的步长都严格劣于 $2^n$ （即其他步长能到达的点集都会被一个步长为 $2^n$ 的射线能到达的点集严格包含）。

那么我们就只需要枚举步长为 $2^n$ 的射线了，对于同侧的点，只要 $mod$ 二倍步长同余就说明在同一射线上，对于异侧的点则需要先加上步长再取模。

由于模数较大而 $n,m$ 很小，可以用 $map$ 来统计。

代码

学习了一波读入姿势，感觉很强。

#include<bits/stdc++.h>
using namespace std;
const int M=1e5+5;
int up[M],down[M],ans=2,n,m;
map<int,int>mp;
void in(){scanf("%d%*d",&n);for(int i=1;i<=n;++i)scanf("%d",&up[i]);scanf("%d%*d",&m);for(int i=1;i<=m;++i)scanf("%d",&down[i]);}
void ac()
{
	for(int d=2;d<=1e9;d<<=1)
	{
		mp.clear();
		for(int i=1;i<=n;++i)++mp[up[i]%d],ans=max(ans,mp[up[i]%d]);
		for(int i=1;i<=m;++i)++mp[(down[i]+(d>>1))%d],ans=max(ans,mp[(down[i]+(d>>1))%d]);
	}
	printf("%d",ans);
}
int main(){in();ac();}