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>
#define min(a,b) (((a)<(b))?(a):(b))
int main(int argc, char const *argv[]){
	long h,w,ans,i,a,b,c;
	scanf("%ld %ld",&h,&w);
	if(h%3==0 || w%3==0){
		printf("0\n");
	}else{
		ans=min(w*(h%3),(w%3)*h);
		for(i=1;i<w;i++){
			a=i*h;
			b=(w-i)*(h/2);
			c=(w-i)*(h/2+h%2);
			if(a<b){
				ans=min(ans,c-a);
			}else{
				if(a>c){
					ans=min(ans,a-b);
				}else{
					ans=min(ans,c-b);
				}
			}
		}
		for(i=1;i<h;i++){
			a=i*w;
			b=(h-i)*(w/2);
			c=(h-i)*(w/2+w%2);
			if(a<b){
				ans=min(ans,c-a);
			}else{
				if(a>c){
					ans=min(ans,a-b);
				}else{
					ans=min(ans,c-b);
				}
			}
		}
		printf("%ld\n",ans);
	}
	return 0;
}