codeforces934 A. A Compatible Pair【暴力】

A. A Compatible Pair

time limit per test1 second

memory limit per test256 megabytes

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.

题意： 给你a,b两个序列，第一个人在a序列选出一个数，第二个人也在a序列中选出一个数，二人不能选一个数，尽量的满足第一个人选出的数乘上序列b的任意一个数尽量小，同时要满足第二人选出的数乘上序列b的任意一数尽量大

分析： 因为里面有负数，所以肯定某个极端，我们枚举下即可，如果a序列只有两个数，要么是第一个要么是第二个，直接枚举即可，要是多个的话，只有六种可能，直接枚举即可

参考代码

#include<bits/stdc++.h>

using namespace std;

#define ll long long
const ll INF_LL = 9223372036854775807LL;

ll a[maxn];
ll b[maxn];

int main(){
ios_base::sync_with_stdio(0);
int n,m;cin>>n>>m;
for(int i = 0;i < n;i++) cin>>a[i];
for(int i = 0;i < m;i++) cin>>b[i];
sort(a,a+n);
sort(b,b+m);
ll a1 = a[0],a2 = a[1],a3 = a[n - 2],a4 = a[n - 1];
ll b1 = b[0],b2 = b[1],b3 = b[m - 2],b4 = b[m - 1];
ll res = -INF_LL;
if(n > 2) {
res = max(res,a2*b1);
res = max(res,a2*b4);
res = max(res,a3*b1);
res = max(res,a3*b4);
res = max(res,min(a[n - 1]*b[m - 1],a[0]*b[0]));
res = max(res,min(a[n - 1]*b[0],a[0]*b[m - 1]));
} else {
res = max(res,a2*b1);
res = max(res,a3*b4);
}
cout<<res<<endl;

return 0;
}

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