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)


// C - Chocolate Bar

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
// const int INF = 2147483647;
const ll INF = 9223372036854775807;
// const ll MOD = 1e9 + 7;

int main() {
	ll H, W;
	cin >> H >> W;

	// 横だけで分割
	ll ans1;
	if (H % 3 == 0) ans1 = 0;
	else ans1 = W;

	// 縦だけで分割
	ll ans2;
	if (W % 3 == 0) ans2 = 0;
	else ans2 = H;

	// 横で分割し、残りを縦で分割
	ll ans3 = INF;
	ll W1, W2;
	if (W % 2 == 0) {
		W1 = W / 2;
		W2 = W1;
	} else {
		W1 = W / 2;
		W2 = W1 + 1;
	}

	for (int i=0; i<=H; i++) {
		ll T = W * i;
		ll S1 = W1 * (H - i);
		ll S2 = W2 * (H - i);

		if (T < S2) {
			ans3 = min(ans3, S2 - T);
		} else {
			ans3 = min(ans3, T - S1);
		}
	}

	// 縦で分割し、残りを横で分割
	ll ans4 = INF;
	ll H1, H2;
	if (H % 2 == 0) {
		H1 = H / 2;
		H2 = H1;
	} else {
		H1 = H / 2;
		H2 = H1 + 1;
	}

	for (int i=0; i<=W; i++) {
		ll T = H * i;
		ll S1 = H1 * (W - i);
		ll S2 = H2 * (W - i);

		if (T < S2) {
			ans4 = min(ans4, S2 - T);
		} else {
			ans4 = min(ans4, T - S1);
		}
	}

	ll ans = min(ans1, min(ans2, min(ans3, ans4)));

	cout << ans << endl;

	return 0;
}