[2020 Winter] [maximum] gmoj2090

Title Description

A given number n, {x1, x2, ..., xn} which selects at least a number of requirements, up to the number n, such that the maximum sum of products.

Entry

The first line integer n, the number represents the number of
the next n lines, each line an integer xi, -10 ≤xi≤ 10

Export

Output line, represents the maximum product

Sample input

Sample Input1:

3
-1
2
-4

Sample Input2:

3
3
2
-4

Sample Output

Sample Output1:

8

Sample Output2:

6

hint

For 70% of the data: 1 ≤ n ≤ 9
to 100% of the data: 1 ≤ n ≤ 18, -10 ≤xi≤ 10

analysis

This question to open long long !!
input, we judge three cases:

1. xi> 0 then use it directly by a sum
2. xi <0 In this case the number of recording a negative We f, b [f] save up to negative, but also with the product of all negative s1 calculated.
3. xi = 0 to skip
   and then we do a sort the array b (small to large)
   if s1> 0, direct output * SUM s1;
   ! If f% 2 = 0, then we ignore the biggest negative number (absolute minimum), take up with the sum.

The Code!

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
long long n,f,s=1,s1=1,a[20];
int main()
{
    freopen("max.in","r",stdin);
	freopen("max.out","w",stdout); 
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
    	int x;
    	scanf("%d",&x);
    	if(n==1)
		{
			cout<<x;
			return 0;
		} 
    	if(x<0)
		{
		    f++;
			a[f]=x;
			s1*=x;
		} 
    	if(x>0) s*=x;
	}
	for(int i=1;i<=f-1;i++)
	   for(int j=i+1;j<=f;j++)
	      if(a[i]>a[j]) swap(a[i],a[j]); 
	if(s1>0&&f%2==0) cout<<s1*s;
	else if(f%2!=0)
	{
		for(int i=1;i<=f-1;i++)
		{
			s*=a[i];
		}
		cout<<s;
	}
	fclose(stdin);
	fclose(stdout);
    return 0;
}