Problem:
Score : 400 points

Problem Statement

There is a bar of chocolate with a height of H blocks and a width of W blocks.
Snuke is dividing this bar into exactly three pieces.
He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle.

Snuke is trying to divide the bar as evenly as possible.
More specifically, he is trying to minimize S_{max} - S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece.
Find the minimum possible value of S_{max} - S_{min}.

Constraints

2 ≤ H, W ≤ 10^5

Input

Input is given from Standard Input in the following format:

H W

Output

Print the minimum possible value of S_{max} - S_{min}.

Sample Input 1

3 5

Sample Output 1

0

In the division below, S_{max} - S_{min} = 5 - 5 = 0.

Sample Input 2

4 5

Sample Output 2

2

In the division below, S_{max} - S_{min} = 8 - 6 = 2.

Sample Input 3

5 5

Sample Output 3

4

In the division below, S_{max} - S_{min} = 10 - 6 = 4.

Sample Input 4

100000 2

Sample Output 4

1

Sample Input 5

100000 100000

Sample Output 5

50000

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>

int main()
{
  int h, w, i, y, y1;
  int x1, x2, x3;
  int min, max;
  long long int answer1 = 1e11;

  scanf("%d %d", &h, &w);
  if(h % 3 == 0 || w % 3 == 0)
  {
    printf("0");
  }
  else
  {
    //1
    for(i=0; i<=h; i++)
    {
      //
      x1 = i * w;
      x2 = (h - i) * (w / 2);
      x3 = (h - i) * (w - w / 2);
      if(x1 <= x2 && x1 <= x3)
      {
        min = x1;
        if(x2 < x3) max = x3;
        else max = x2;
      }
      else if(x2 <= x1 && x2 <= x3)
      {
        min = x2;
        if(x1 < x3) max = x3;
        else max = x1;
      }
      else
      {
        min = x3;
        if(x1 < x2) max = x2;
        else max = x1;
      }

      if(answer1 > max - min)
      {
        answer1 = max - min;
      }
      //
      y1 = (h - i) / 2;
      x2 = y1 * w;
      x3 = (h - i - y1) * w;
      if(x1 <= x2 && x1 <= x3)
      {
        min = x1;
        if(x2 < x3) max = x3;
        else max = x2;
      }
      else if(x2 <= x1 && x2 <= x3)
      {
        min = x2;
        if(x1 < x3) max = x3;
        else max = x1;
      }
      else
      {
        min = x3;
        if(x1 < x2) max = x2;
        else max = x1;
      }
      //
      if(answer1 > max - min)
      {
        answer1 = max - min;
      }
    }
    //2
    y = h;
    h = w;
    w = y;
    for(i=0; i<=h; i++)
    {
      //
      x1 = i * w;
      x2 = (h - i) * (w / 2);
      x3 = (h - i) * (w - w / 2);
      if(x1 <= x2 && x1 <= x3)
      {
        min = x1;
        if(x2 < x3) max = x3;
        else max = x2;
      }
      else if(x2 <= x1 && x2 <= x3)
      {
        min = x2;
        if(x1 < x3) max = x3;
        else max = x1;
      }
      else
      {
        min = x3;
        if(x1 < x2) max = x2;
        else max = x1;
      }
      //
      if(answer1 > max - min)
      {
        answer1 = max - min;
      }
      //
      y1 = (h - i) / 2;
      x2 = y1 * w;
      x3 = (h - i - y1) * w;
      if(x1 <= x2 && x1 <= x3)
      {
        min = x1;
        if(x2 < x3) max = x3;
        else max = x2;
      }
      else if(x2 <= x1 && x2 <= x3)
      {
        min = x2;
        if(x1 < x3) max = x3;
        else max = x1;
      }
      else
      {
        min = x3;
        if(x1 < x2) max = x2;
        else max = x1;
      }
      //
      if(answer1 > max - min)
      {
        answer1 = max - min;
      }
    }
    printf("%lld", answer1);
  }
  return 0;
}