https://www.geeksforgeeks.org/minimum-product-subset-array/

Given an array a, we have to find minimum product possible with the subset of elements present in the array. The minimum product can be single element also.

Examples:

Input : a[] = { -1, -1, -2, 4, 3 }
Output : -24
Explanation : Minimum product will be ( -2 -1 -1 4 3 ) = -24

Input : a[] = { -1, 0 }
Output : -1
Explanation : -1(single element) is minimum product possible

Input : a[] = { 0, 0, 0 }
Output : 0

  1. 如果有偶数个负数且没有零，则结果为除最大负数之外的所有值的乘积。
  2. 如果有奇数个负数而没有零，那么结果就是所有乘积。
  3. 如果只有零和正，没有负，则结果为0。例外情况是，当没有负数且所有其他元素为正时，我们的结果应为第一个最小正数
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#include <bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
int max_neg=-INF,min_pos=INF,count_neg=0,count_zero=0,prod=1,t,n;
int main(){
	freopen("in","r",stdin);
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&t);
		if(t==0) {
			count_zero++;
			continue;
		}
		if(t<0) {
			count_neg++; 
			max_neg = max(max_neg, t); 
		}
		if(t>0)	min_pos = min(min_pos, t);  
		prod*=t;
	}
    if (count_zero == n ||(count_neg == 0 && count_zero > 0)) 
        printf("0");
    else if (count_neg == 0) 
        printf("%d",min_pos);
    else if (!(count_neg & 1) && count_neg != 0) 
    	printf("%d",prod / max_neg);
    else printf("%d",prod);
	return 0;
}