Traffic （難度不夠，題意來湊？）

傳送門

Avin is observing the cars at a crossroads. He finds that there are n cars running in the east-west direction with the i-th car passing the intersection at time ai . There are another m cars running in the north-south direction with the i-th car passing the intersection at time bi . If two cars passing the intersections at the same time, a traffic crash occurs. In order to achieve world peace and harmony, all the cars running in the north-south direction wait the same amount of integral time so that no two cars bump. You are asked the minimum waiting time.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 1, 000). The second line contains n distinct integers ai (1 ≤ ai ≤ 1, 000). The third line contains m distinct integers bi (1 ≤ bi ≤ 1, 000).

Output

Print a non-negative integer denoting the minimum waiting time.

Sample Input

1 1
1
1
1 2
2
1 3

Sample Output

1
0

題解：

做不出這道題的童鞋相信都卡在了這個坑裏。

all the cars running in the north-south direction wait the same amount of integral time so that no two cars bump.

去掉定語後是：all the cars wait the same time. 所有汽車等待時間相同。

這句話的意思是：所有南北方向上的汽車要等待相同的時間，以至於沒有兩輛汽車相撞。

這句話的關鍵是所有汽車等待時間相同，要麼全都是5min，要麼全都是10min，不能這輛汽車等1min、那輛汽車等2min的。

暴力枚舉即可。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdio>

using namespace std;

typedef long long ll;

const int maxn=1e4+10;

int a[maxn],b[maxn],dis[maxn];

int main()
{	
	int n,m;
	while(cin>>n>>m)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(dis,0,sizeof(dis));
		
		for(int i=0;i<n;i++) 
		{
			cin>>a[i];
			dis[a[i]]=1;
		}
		for(int i=0;i<m;i++) cin>>b[i];
		int ans;
		for(int i=0;i<1001;i++)
		{
			int flag=0;
			for(int j=0;j<m;j++)
			{
				if(dis[b[j]+i])
				{
					flag=1;
					break;
				}
			}
			if(!flag)
			{
				ans=i;
				break;
			}
		}
		cout<<ans<<endl;
	}
	
	return 0;
}