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: Wrong Answer)

#include <stdio.h>
#include<stdlib.h>
int main(){
    long long int a,b,c,d,product=0;
    scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
  if(a==0||b==0||c==0||d==0)
  {
   if(a>=0&&c>=0) product=b*d;
  else if(a<0&&b>0&&c>0) product=b*d;
  else if(a>0&&c<0&&d>0) product=b*d;
  else if(a<0&&b<0&&c>0) product=b*c;
  else if(a>0&&c<0&&d<0) product=a*d;
  else if(a<0&&b<0&&c<0&&d<0)product=a*b;
  else if(a<0&&b>0&&c<0&&d>0){
      if(abs(a*c)>=abs(b*d))product=a*c;
    else product=b*d;
  }
    if(product<0)product=0;
  printf("%d",product);
}
else{
  if(a>=0&&c>=0) product=b*d;
  else if(a<0&&b>0&&c>0) product=b*d;
  else if(a>0&&c<0&&d>0) product=b*d;
  else if(a<0&&b<0&&c>0) product=b*c;
  else if(a>0&&c<0&&d<0) product=a*d;
  else if(a<0&&b<0&&c<0&&d<0)product=a*b;
  else if(a<0&&b>0&&c<0&&d>0){
      if(abs(a*c)>=abs(b*d))product=a*c;
    else product=b*d;
  }
  printf("%d",product);
}
  return 0;
}