codeforces 559A Gerald's Hexagon

最新推荐文章于 2019-11-13 19:18:50 发布

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本文介绍了一种算法，用于计算特殊六边形内部由平行于六边形边绘制的线条所形成的1厘米边长的等边三角形的数量。通过按层计算的方法，实现了对不同形状六边形的有效处理。

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Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to $\text{[math]}$ . Then he measured the length of its sides, and found that each of them is equal to an integer number of centimeters. There the properties of the hexagon ended and Gerald decided to draw on it.

He painted a few lines, parallel to the sides of the hexagon. The lines split the hexagon into regular triangles with sides of 1 centimeter. Now Gerald wonders how many triangles he has got. But there were so many of them that Gerald lost the track of his counting. Help the boy count the triangles.

Input

The first and the single line of the input contains 6 space-separated integersa₁, a₂, a₃, a₄, a₅ anda₆ (1 ≤ a_i ≤ 1000) — the lengths of the sides of the hexagons in centimeters in the clockwise order. It is guaranteed that the hexagon with the indicated properties and the exactly such sides exists.

Output

Print a single integer — the number of triangles with the sides of one 1 centimeter, into which the hexagon is split.

Examples

Input

1 1 1 1 1 1

Output

Input

1 2 1 2 1 2

Output

Note

This is what Gerald's hexagon looks like in the first sample:

$\text{[math]}$

And that's what it looks like in the second sample:

$\text{[math]}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

计算几何~

刚开始以为是正六边形白白浪费了好多时间……

因为所有角度都相等，所以对边一定平行。这样我们就能直接按“层”来做这道题。

我们以1、4两条边为顶和底，剩下的是边。

“层”一共有三种情况：两边都没有到转弯点（下一层比现在这层大1），一边过了转弯点（此时两边平行，下一层和现在一层相等），两边都过了转弯点（与第一种情况相反，下一层比这一层小1）。暴力按层计算就好了~

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int n,m,a[7],l,r,mid,mid2,now,ans;
bool flag1,flag2;

int main()
{
	for(int i=1;i<=6;i++) scanf("%d",&a[i]);now=a[1];
	while(now!=a[4] || !flag1 || !flag2)
	{
		l++;
		if(l>a[6] && !flag1) flag1=1;
		r++;
		if(r>a[2] && !flag2) flag2=1;
		if(flag1+flag2==0) ans+=now+now+1;
		else if(flag1+flag2==2) ans+=now+now-1;
		else ans+=now*2;
		if(flag1+flag2==0) now++;
		else if(flag1+flag2==2) now--;
	}
	printf("%d\n",ans);
	return 0;
}