題目：

Description
Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to 120°. 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 Input
Input
1 1 1 1 1 1
Output
6
Input
1 2 1 2 1 2
Output
13
Hint
This is what Gerald's hexagon looks like in the first sample:

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

題目的意思是給你一個(gè)等角六邊形（六個(gè)角都為120°），六條邊的長(zhǎng)度都給你，按與邊平行劃線蔚润， 問(wèn)最多會(huì)分成多少個(gè)邊長(zhǎng)為1的小三角形乙各。

20161103_161028.jpg

根據(jù)等角六邊形的特點(diǎn)（六個(gè)角都為120°工禾，對(duì)邊平行）赌蔑，可以將該六邊形劃分成如圖所示的情況悯周。求出六邊形的面積拌蜘，再除以小三角形的面積杆烁，即可得到最終的結(jié)果。

參考代碼：

#include <iostream>
#include <algorithm>
#include <cmath>
#define PI acos(-1.0) 
using namespace std;
const int N = 6+2;

int a[N];

void input() {
    for (int i = 1;i <= 6;++i) {
        cin >> a[i];
    }
}

int result() {
    double tixing = (a[4] + a[1]) * ((a[3] * sqrt(3.0) / 2.0) + a[2] * (sqrt(3.0) / 2.0)) / 2.0;
    double sanjiao1 = a[2] * a[3] * (sqrt(3.0) / 2.0) / 2.0;
    double sanjiao2 = a[5] * a[6] * (sqrt(3.0) / 2.0) / 2.0;
    double total = sanjiao1 + sanjiao2 + tixing;
    double sanjiao = 1 * 1 * (sqrt(3.0) / 2.0) / 2.0;
    //cout << (total / sanjiao) << endl;
    int ans = (int) (total / sanjiao + 0.5);
    return ans;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    input();
    int ans = result();
    cout << ans << endl;
    return 0;
}