ACM----CodeForces Round #670 (Div. 2)B. Maximum Product

You are given an array of integers 𝑎1,𝑎2,…,𝑎𝑛. Find the maximum possible value of 𝑎𝑖𝑎𝑗𝑎𝑘𝑎𝑙𝑎𝑡 among all five indices (𝑖,𝑗,𝑘,𝑙,𝑡) (𝑖<𝑗<𝑘<𝑙<𝑡).

Input
The input consists of multiple test cases. The first line contains an integer 𝑡 (1≤𝑡≤2⋅104) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer 𝑛 (5≤𝑛≤105) — the size of the array.

The second line of each test case contains 𝑛 integers 𝑎1,𝑎2,…,𝑎𝑛 (−3×103≤𝑎𝑖≤3×103) — given array.

It’s guaranteed that the sum of 𝑛 over all test cases does not exceed 2⋅105.

Output
For each test case, print one integer — the answer to the problem.

Example
inputCopy
4
5
-1 -2 -3 -4 -5
6
-1 -2 -3 1 2 -1
6
-1 0 0 0 -1 -1
6
-9 -7 -5 -3 -2 1
outputCopy
-120
12
0
945
Note
In the first test case, choosing 𝑎1,𝑎2,𝑎3,𝑎4,𝑎5 is a best choice: (−1)⋅(−2)⋅(−3)⋅(−4)⋅(−5)=−120.

In the second test case, choosing 𝑎1,𝑎2,𝑎3,𝑎5,𝑎6 is a best choice: (−1)⋅(−2)⋅(−3)⋅2⋅(−1)=12.

In the third test case, choosing 𝑎1,𝑎2,𝑎3,𝑎4,𝑎5 is a best choice: (−1)⋅0⋅0⋅0⋅(−1)=0.

In the fourth test case, choosing 𝑎1,𝑎2,𝑎3,𝑎4,𝑎6 is a best choice: (−9)⋅(−7)⋅(−5)⋅(−3)⋅1=945.

题目大意：在给定序列中选五个数使得乘积最大。

解题思路：先将给定序列进行排序，排序后求序列左侧五个数乘积，左侧四个数和右侧一个数乘积，左侧三个数和右侧两个数乘积..........最后求右侧五个数的乘积，在这六个数中取最大值。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main(){
    int n,m,f;
    long long int a[100000],s,result;
    cin >> n;
    while(n--){
        cin >> m;
        for(int i=0;i<m;i++)
            cin >> a[i];
        sort(a,a+m);
        result=a[0]*a[1]*a[2]*a[3]*a[4];   
        for(int i=4;i>=0;i--){
            s=1;
            for(int j=0;j<i;j++)   
                s*=a[j];
            for(int j=0;j<5-i;j++)    
                s*=a[m-1-j];
            result=max(result,s);     
        }
        cout <<result<< endl;
    }
    return 0;
}