【打CF，学算法——三星级】Codeforces Round #313 (Div. 2) C. Gerald's Hexagon

【CF简单介绍】

提交链接：http://codeforces.com/contest/560/problem/C

题面：

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 integersa₁, a₂, a₃, a₄, a₅
anda₆ (1 ≤ a_i ≤ 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

解题：

画着画着发现。每个六边形能够用相邻两条边构成的平行四边形切割，有些情况恰能够切割，有些不可，中间存在一个三角形。

不能够的是，相隔2条的边的边长之差不为0。

然后依据这个差。就能够算出中间还有几个未切割的三角形。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int main()
{
    int a,b,c,d,e,f,sum,ans;
    sum=ans=0;
    scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f);
    ans+=2*(a*b+c*d+e*f);
    sum=abs(a-d);
    sum=sum*sum;
    ans+=sum;
    printf("%d\n",ans);
    return 0;
}