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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.
1 1 1 1 1 1
6
1 2 1 2 1 2
13
地址：http://codeforces.com/contest/560/problem/C
思路：额，这道题想了挺久，刚开始，想画图找规律，分层次找了半天，得出一些特殊情况的规律，但是没有什么卵用。
然后想到了面积，想到的有点迟，然后就敲代码了。结果对6边形的理解有错误，最后错了，于是乎，百度了。
数学太差没办法。。
感觉上有两种做法：
1： 通过面积求
直接求面积：http://blog.csdn.net/winddreams/article/details/47016541
间接求面积：http://blog.csdn.net/crescent__moon/article/details/47017283
我的代码是直接求面积
#include <cstdio> using namespace std; int main(){ int edge[6]; int t_area = 1; int h_area; int cnt; while(scanf("%d", &edge[0]) != EOF){ for(int i = 1; i < 6; i ++){ scanf("%d", &edge[i]); } h_area = (edge[0] + edge[3]) * (edge[1] + edge[2]) + edge[1] * edge[2] + edge[4] * edge[5]; cnt = h_area / t_area; printf("%d\n", cnt); } return 0; }
第2种应该就是找规律了：
时间比较晚了第2天再想想 然后再补。
2015-07-27 23:47:39
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原文地址：http://www.cnblogs.com/0-happy-0/p/4681616.html
