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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 ≤ 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.
Print a single integer — the number of triangles with the sides of one 1 centimeter, into which the hexagon is split.
题目大意：给出一个等角六边形的六条边长(都为整数厘米)，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
六边形的全部都被分为了小三角形，所以用六边形的面积/三角形的面积就是三角形的数量
六边形的面积：设六条边长为a[1]到a[6]
等角六边形的对边平行，连接两个对边的定点，就得到一个梯形和两个三角形，梯形的上底是a[4]，下底是a[1]，高由划红线的地方算出a[3]*sin(PI/3.0)+a[2]*sin(PI/3.0)，这样就可以算出梯形的面积，其余的两个三角形的面积为a[2]*a[3]*sin(PI/3.0)/2， a[5]*a[6]*sin(PI/3.0)/2
小三角形的面积：1*1*sin(PI/3.0)/2.0 ;
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
using namespace std ;
#define LL __int64
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
double a[7] ;
int main() {
int i ;
double l , s ;
for(i = 1 ; i <= 6 ; i++)
scanf("%lf", &a[i]) ;
s = (a[1]+a[4])*(a[2]+a[3]) + a[2]*a[3]+ a[5]*a[6];
printf("%d\n", (int)(s+0.5)) ;
return 0 ;
}
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