2015 HUAS Summer Trainning #6~O

Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to . 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.

The first and the single line of the input contains 6 space-separated integers a₁, a₂, a₃, a₄, a₅ and a₆ (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.

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

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

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

解题思路：题目的意思是输入六个整数组成一个六边形确定其能组成多少个边长为1的三角形。这里需要注意的是它的输入是按照顺时针输入的，所以可以顺时针将它们组成六边形。我觉得可以将其补成一个大的等边三角形，然后将大三角形边长乘以边长减去补上的三个小三角形的边长乘以边长之和即可。

程序代码：

#include<stdio.h>
#include<algorithm>
using namespace std;

int main()
{
	int a1,a2,a3,a4,a5,a6;
	scanf("%d%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5,&a6);
	int t=a1+a2+a3;
	int count=t*t-(a1*a1+a3*a3+a5*a5);
	printf("%d\n",count);
	return 0;
}