CodeForces 559A Gerald's Hexagon

题目：

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:

题目的意思是给你一个等角六边形（六个角都为120°），六条边的长度都给你，按与边平行划线， 问最多会分成多少个边长为1的小三角形。

20161103_161028.jpg

根据等角六边形的特点（六个角都为120°，对边平行），可以将该六边形划分成如图所示的情况。求出六边形的面积，再除以小三角形的面积，即可得到最终的结果。

参考代码：

#include 
#include 
#include 
#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;
}