G - Traffic

vin 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

题意：n辆车从一边经过十字路口，m辆车从另一边经过十字路口，当n方向的车经过路口时，m方向的车必须停下等待。给定车辆经过路口的时间，问m方向的车最少要等多久

题解：假设等待时间时t,当两个方向 的车同时经过路口时需要等，即b[j]+t==a[i],枚举输出最大的t即可

#include<iostream>

#include<string.h>

using namespace std;

int a[],b[];

int main()

{

    int n,m,x;

    while(cin>>n>>m)

    {

        memset(a,,sizeof(a));

        memset(b,,sizeof(b));

        for(int i=;i<n;i++)

        {

            cin>>x;

            a[x]=;

        }

        for(int i=;i<m;i++)

          cin>>b[i];

        int t=;

        for(int i=;i<m;i++)

        {

            if(a[b[i]+t])//如果b[i]+t==a[j]

            {

                t++;

                i=-;

            }

        }

        cout<<t<<endl;

    }

    return ;

}

巴特西

G - Traffic

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