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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "vector"
#include "iostream"
#include "string"
#include "algorithm"

using namespace std;

int main()
{
	long long h, w;
	cin >> h >> w;
	long long r = (h * w) % 3;
	if (r == 0)
	{
		cout << 0 << endl;
		return 0;
	}
	long long rt = 1e12;
	for (int i = 1; i <= w; i++)
	{
		long long t1 = h * i;
		long long temp = h * w - t1;
		long long t2;
		long long t3 = (temp / h) * (h / 2);
		if (h % 2 == 0)
		{
			t2 = (temp / h) * (h / 2);
		}
		else {
			t2 = (temp / h) * ((h / 2) + 1);
		}
		long long t = 0;
		if (t1 > t2)
		{
			t = t1 - t3;
		}
		else {
			if (t3 > t1)
			{
				t = t2 - t1;
			}
			else {
				t = t2 - t3;
			}
		}
		if (t > rt)
		{
			break;
		}
		rt = t;
	}
	long long rt2 = 1e12;
	for (int i = 1; i <= h; i++)
	{
		long long t1 = w * i;
		long long temp = h * w - t1;
		long long t2;
		long long t3 = (temp / w) * (w / 2);
		if (w % 2 == 0)
		{
			t2 = (temp / w) * (w / 2);
		}
		else {
			t2 = (temp / w) * ((w / 2) + 1);
		}
		long long t = 0;
		if (t1 > t2)
		{
			t = t1 - t3;
		}
		else {
			if (t3 > t1)
			{
				t = t2 - t1;
			}else {
				t = t2 - t3;
			}

		}
		if (t > rt2)
		{
			break;
		}
		rt2 = t;
	}
	long long rt3 = 1e12;
	for (int i = 1; i <= w; i++)
	{
		long long t1 = h * i;
		long long temp = h * w - t1;
		long long t2, t3;
		t3 = (temp / h) * (h / 2);
		if (h % 2 == 0)
		{
			t2 = (temp / h) * (h / 2);
		}
		else {
			t2 = (temp / h) * ((h / 2) + 1);
		}
		long long t = 0;
		if (t1 > t2)
		{
			t = t1 - t3;
		}
		else {
			if (t3 > t1)
			{
				t = t2 - t1;
			}else {
				t = t2 - t3;
			}

		}
		if (t > rt3)
		{
			break;
		}
		rt3 = t;

	}
	long long rt4 = 1e12;
	for (int i = 1; i <= h; i++)
	{
		long long t1 = w * i;
		long long temp = h * w - t1;
		long long t2, t3;
		t3 = (temp / w) * (w / 2);
		if (w % 2 == 0)
		{
			t2 = (temp / w) * (w / 2);
		}
		else {
			t2 = (temp / w) * ((w / 2) + 1);
		}
		long long t = 0;
		if (t1 > t2)
		{
			t = t1 - t3;
		}
		else {
			if (t3 > t1)
			{
				t = t2 - t1;
			}else {
				t = t2 - t3;
			}

		}
		if (t > rt4)
		{	
			break;
		}
		rt4 = t;
	}
	cout << min({ rt, rt2, rt3, rt4 }) << endl;
	return 0;
}