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

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:

很有意思的一道题目：

给你六边形的六个边长，求出该六边形内部有多少个单位三角形。

分析：

第一个图还看不出啥来，关键第二个图，已经非常接近三角形了，所以把这个三角形给补出来，相当于找三角形内部的单位三角形，减去三个角即可。

三角形找规律肯定比六边形好找吧。

三角形边长为1时，个数是1 = 1 * 1；

三角形边长为2时，个数是4 = 2 * 2 ；

三角形边长为3时，个数是9 = 3 * 3；

三角形边长为4时，个数是16 = 4 * 4；

规律很清晰了，个数 = 边长的平方；

很容易求出三角形边长是a1 + a2 + a3；

所以三角个数是(a1 + a2 + a3) ^2 - a1^2 - a3 ^2 - a5^2;

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    ll a1,a2,a3,a4,a5,a6;
    while(cin >> a1 >> a2 >> a3 >> a4 >> a5 >> a6){
        ll all = (a1+a2+a3)*(a1+a2+a3)- a1*a1-a3*a3-a5*a5;
        cout << all << endl;
    }
    return 0;
}