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<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long ll;

//hが奇数でwを動かすとき
int odd(int h, int w, int x,int m) {
	if (h*x >= (h + 1)*(w - x) / 2)return min(m, h*x - (h - 1)*(w - x) / 2);
	if (h*x >= (h - 1)*(w - x) / 2)return min(m, w - x);
	if (h*x < (h - 1)*(w - x) / 2)return min(m, -h*x + (h + 1)*(w - x) / 2);
}

int main(void) {
	int h, w, ans = 10000000;
	cin >> h >> w;
	if (h % 3 == 0 || w % 3 == 0)cout << 0 << endl;
	else {
		if (h % 2 == 0 && w % 2 == 0) {
			for (int i = 1;i < h;i++)ans = min(ans, abs(w*i - w / 2 * (h - i)));
			for (int i = 1;i < w;i++)ans = min(ans, abs(h*i - h / 2 * (w - i)));
			cout << ans << endl;
		}
		else {
			if (h % 2 == 0) {
				for (int i = 1;i < w;i++)ans = min(ans, abs(h*i - h / 2 * (w - i)));
				for (int i = 1;i < h;i++)ans = min(ans, odd(w, h, i, ans));
				cout << ans << endl;
			}
			else {
				if (w % 2 == 0) {
					for (int i = 1;i < h;i++)ans = min(ans, abs(w*i - w / 2 * (h - i)));
					for (int i = 1;i < w;i++)ans = min(ans, odd(h, w, i, ans));
					cout << ans << endl;
				}
				else {
					for (int i = 1;i < h;i++)ans = min(ans, odd(w, h, i, ans));
					for (int i = 1;i < w;i++)ans = min(ans, odd(h, w, i, ans));
					cout << ans << endl;
				}
			}
		}
	}
	return 0;
}