Codeforces Round #462 (Div. 2) A Compatible Pair

A. A Compatible Pair

time limit per test1 second

memory limit per test256 megabytes

Problem Description

Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.

Little Tommy has n lanterns and Big Banban has m lanterns. Tommy’s lanterns have brightness a1, a2, …, an, and Banban’s have brightness b1, b2, …, bm respectively.

Tommy intends to hide one of his lanterns, then Banban picks one of Tommy’s non-hidden lanterns and one of his own lanterns to form a pair. The pair’s brightness will be the product of the brightness of two lanterns.

Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.

You are asked to find the brightness of the chosen pair if both of them choose optimally.

Input

The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).

The second line contains n space-separated integers a1, a2, …, an.

The third line contains m space-separated integers b1, b2, …, bm.

All the integers range from - 109 to 109.

Output

Print a single integer — the brightness of the chosen pair.

Examples

input

2 2

20 18

2 14

output

252

input

5 3

-1 0 1 2 3

-1 0 1

output

2

Note

In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.

In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.

解题心得：

题意很简单就是给你两个数列a和b，要求你在a中选一个数b中选一个数两数的乘积最大，但是a为了使乘积尽量小会删去一个数，要求你输出除a删去了那个数之后能够的到的乘积最大的那个数。
数据范围很小也就50，可以先找出乘积最大的z中的那个数，然后标记，在找标记之后最大的就行了。暴力最稳妥。

代码写得很烂，切勿模仿

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 55;

ll n,m;

ll a[maxn],b[maxn];

map <ll,ll> maps;

struct NODE{

    ll x,y,va;

    friend bool operator < (const NODE a,const NODE b){

        return a.va < b.va;

    }

};

int main() {

    priority_queue <NODE> qu;

    scanf("%lld%lld", &n, &m);

    for (int i = 0; i < n; i++)

        scanf("%lld", &a[i]);

    for (int i = 0; i < m; i++)

        scanf("%lld", &b[i]);

    for(int i=0;i<n;i++)

        for(int j=0;j<m;j++){

            NODE temp;

            long long c = a[i]*b[j];

            temp.x = i;

            temp.y = j;

            temp.va = c;

            qu.push(temp);

        }

    bool flag = false;

    while(!qu.empty()){

        NODE temp = qu.top();

        qu.pop();

        if(!flag){

            maps[temp.x] = 233;

            flag = true;

        }

        if(maps[temp.x] != 233){

            printf("%lld",temp.va);

            break;

        }

    }

    return 0;

}