Problem:
Score : 200 points

Problem Statement

Given are integers a,b,c and d.
If x and y are integers and a \leq x \leq b and c\leq y \leq d hold, what is the maximum possible value of x \times y?

Constraints

-10^9 \leq a \leq b \leq 10^9

-10^9 \leq c \leq d \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

a b c d

Output

Print the answer.

Sample Input 1

1 2 1 1

Sample Output 1

2

If x = 1 and y = 1 then x \times y = 1.
If x = 2 and y = 1 then x \times y = 2.
Therefore, the answer is 2.

Sample Input 2

3 5 -4 -2

Sample Output 2

-6

The answer can be negative.

Sample Input 3

-1000000000 0 -1000000000 0

Sample Output 3

1000000000000000000

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main(){
  ll t;
  //cin>>t;
  //while(t--){
    ll a,b=1,c=1,d=0,e=0,f=0,g=0,h=0,e1=0,f1=0;
    cin>>a>>b>>c>>d;
    if(a*b<=0 && c*d<=0){
       e=min(a,b);e1=max(a,b);
        f=min(c,d);f1=max(c,d);
        g=max((e*f),(e1*f1));
        cout<<g<<endl;
    }
    else if(a*b>0 && c*d>0){
       if((a>0 && c<0) || (a<0 && c>0)){
        if(a>0){
            e=min(a,b);
            f=max(c,d);
            cout<<e*f<<endl;
        }
        else{
            e=max(a,b);
            f=min(c,d);
            cout<<e*f<<endl;
        }
       }
       else{
         e=min(a,b);e1=max(a,b);
        f=min(c,d);f1=max(c,d);
        g=max((e*f),(e1*f1));
        cout<<g<<endl;
       }
    }
    else{
            if(a*b>0){
        if(a>0){
        e=max(a,b);
        f=max(c,d);
        cout<<e*f<<endl;
        }
        else{
            e=min(a,b);
            f=min(c,d);
            cout<<e*f<<endl;
        }
            }
            if(c*d>0){
              if(c>0){
                  e=max(a,b);
        f=max(c,d);
        cout<<e*f<<endl;
              }
              else{
               e=min(a,b);
            f=min(c,d);
            cout<<e*f<<endl;
              }
            }

    }

  //}
  }