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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int main()
{
	ll a,b,c,d;cin>>a>>b>>c>>d;
	ll ans1=b*d;ll ans2=b*d;
	ll x=b;ll y=d;
	ll ans3=a*c;
	while(x>a)
	{
		x--;
		ans1=max(ans1,x*d);                   // a<x<b; c<y<d
	}
	ans1=max(ans1,ans3);
	while(y>c)
	{
		y--;
		ans2=max(ans2,y*b);
	}
	ans2=max(ans2,ans3);
	cout<<max(ans1,ans2);
}