[2018-4-8]BNUZ套题比赛div2 CodeForces 934A【补】

A. A Compatible Pairtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputNian 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.InputThe 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.OutputPrint a single integer — the brightness of the chosen pair.ExamplesinputCopy

2 2
20 18
2 14

outputCopy

252

inputCopy

5 3
-1 0 1 2 3
-1 0 1

outputCopy

2

NoteIn 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：Banban将会从Tommy的数组中取走一个数，和自己数组的一个数组成乘积最大。但是Tommy希望Banban最终组成的数尽可能小，而且它将会提前从自己数组中取走一个数以此来达到目的。输出最终组成的数。题解：只要暴力跑就行了， 第一次将 相乘 最大的数的位置记录下来， 第二次跳过这个位置就可以了。ps：比较的的是 乘积 因为ai 和 bi 都有可能是负数，不能直接排序a数组和b数组。AC代码：#include <bits/stdc++.h>
#define ll long long
#define INF LLONG_MAX

using namespace std;

int main() {
ll n, m, a[55], b[55];
cin >> n >> m;
for (ll i = 0; i < n; i++)
cin >> a[i];
for (ll j = 0; j < m; j++)
cin >> b[j];
ll maxx = -INF, index = -1;
for (ll i = 0; i < n; i++)
for (ll j = 0; j < m; j++)
if (a[i] * b[j] > maxx) {
index = i;
maxx = a[i] * b[j];
}
maxx = -INF;
for (ll i = 0; i < n; i++)
for (ll j = 0; j < m; j++) {
if (i == index)
continue;
maxx = max(a[i] * b[j], maxx);
}
cout << maxx << endl;
}