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 <vector>

using namespace std;

long long int H, W;

long long int f_ans(long long int x, long long int y) {
    if(x%3==0 || y%3==0) {
        return 0;
    }

    long long int h=min(x, y), w=max(x, y);
    long long int Smax=(w/3+1)*H, Smin=0;
    if(w%3==1) {
        long long int tmpmax, tmpmin;
        for(int i=0; i<2; i++) {
            vector<long long int> S(3);
            S[0] = (w/3+i)*h;
            if(h%2==0) {
                S[1] = (H*W - S[0])/2;
            } else {
                if(w-w/3>h) {
                    S[1] = ((w-w/3-i)/2+1)*h;
                } else {
                    S[1] = (h/2+1)*(w-w/3-i);
                }
            }
            S[2] = H*W - S[0] - S[1];
            
            sort(S.begin(), S.end());
            tmpmin = S[0];
            tmpmax = S[2];
            // cout << tmpmin << ", " << tmpmax << endl;
            if(Smax-Smin>tmpmax-tmpmin) {
                Smax = tmpmax;
                Smin = tmpmin;
            }
        }
    } else {
        if(h%2==0) {
            Smin = ((H*W)-Smax)/2;
        } else {
            if(w-(w/3+1)>h) {
                Smin = (w-(w/3+1))/2*h;
            } else {
                Smin = h/2*(w-(w/3+1));
            }
        }
    }
    
    return Smax-Smin;
}

int main() {
    cin >> H >> W;
    cout << f_ans(H, W) << endl;

    return 0;
    
}