Codeforces Round #313 (Div. 2) C. Gerald's Hexagon 几何

C. Gerald's Hexagon

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

The first and the single line of the input contains 6 space-separated integers a1, a2, a3, a4, a5 and a6 (1 ≤ ai ≤ 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.

Sample test(s)

input

1 1 1 1 1 1

output

6

input

1 2 1 2 1 2

output

13

Note

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

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

题意是，给个六角形的六条边，要求里面可以分割成几个三角形，六角形六个角都是120度。直接求总面积再除三角形面积就可以了。 

如图面积为可由黄色的矩形减去四个三角形就可以了。

int main()
{
    while(S(a[0]) != EOF)
    {
        for(int i =1;i<=5;i++) S(a[i]);
        int s = (((a[5] + a[1] + 2 * a[0])) * (a[1] + a[2]) * 2.0
                                    - a[1] * a[1] - a[2] * a[2] - a[4] * a[4] - a[5] * a[5])/2;
        printf("%d\n",s);
    }
    return 0;
}

第二种方法，如图

把一个六角形补成一个等边三角形，我们可以发现任何一个等边三角形可以分割成边长的平方个小的三角形，所以用大的三角形减去三个小的三角形的能形成的三角形即可得到答案，大三角形边长 a[0] + a[1] + a[2] 小三角形边长分别为 a[0] a[2] a[4],所以得到公式

（ a[0] + a[1] + a[2]） *（ a[0] + a[1] + a[2]）－　a[0] * a[0] - a[2] * a[2] - a[4] * a[4]就可以了。