A. Gerald's Hexagon

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/559/problem/A

Description

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 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

1 1 1 1 1 1

Sample Output

6

HINT

题意

给你一个角都是120度的六边形,然后问你里面有多少个边长为1的正三角形

题解:

简单分析一下,我们可以把这个三角形,扩展成一个大的正三角形,然后减去边上的小三角形就好了~

代码

#include <iostream>
#include <cstdio> using namespace std; int a,b,c,d,e,f; int main(){
scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f);
int t=a+b+f;
cout<<t*t-b*b-d*d-f*f<<endl;
return ;
}

Codeforces Round #313 (Div. 1) A. Gerald's Hexagon 数学题的更多相关文章

  1. Codeforces Round #313 (Div. 1) A. Gerald's Hexagon

    Gerald's Hexagon Problem's Link: http://codeforces.com/contest/559/problem/A Mean: 按顺时针顺序给出一个六边形的各边长 ...

  2. Codeforces Round #313 (Div. 2) C. Gerald's Hexagon 数学

    C. Gerald's Hexagon Time Limit: 2 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/559/pr ...

  3. Codeforces Round #313 (Div. 2) C. Gerald's Hexagon

    C. Gerald's Hexagon time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  4. 【打CF,学算法——三星级】Codeforces Round #313 (Div. 2) C. Gerald&#39;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 tes ...

  5. Codeforces Round #313 (Div. 2) C. Gerald&#39;s Hexagon(补大三角形)

    C. Gerald's Hexagon time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  6. Codeforces Round #313 (Div. 2) 560C Gerald&#39;s Hexagon(脑洞)

    C. Gerald's Hexagon time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  7. dp - Codeforces Round #313 (Div. 1) C. Gerald and Giant Chess

    Gerald and Giant Chess Problem's Link: http://codeforces.com/contest/559/problem/C Mean: 一个n*m的网格,让你 ...

  8. Codeforces Round #313 (Div. 2) B. Gerald is into Art 水题

    B. Gerald is into Art Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/560 ...

  9. Codeforces Round #313 (Div. 1) C. Gerald and Giant Chess DP

    C. Gerald and Giant Chess Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest ...

随机推荐

  1. java解析properties文件

    在自动化测试过程中,经常会有一些公用的属性要配置,以便后面给脚本使用,我们可以选择xml, excel或者json格式来存贮这些数据,但其实java本身就提供了properties类来处理proper ...

  2. DouNet学习_收发邮件

    一.收发邮件 --->第一步:发邮件首先要有发送者的邮箱地址和登录的密码才能发送 这些都写在APP里  不要写死 --->第二步:发邮件就要有网络,要添加net.Mail命名空间 要发送的 ...

  3. 项目 erlang启动时死循环

    机子里的otp是新装的 看了一下main 是在util:ensure_started一堆app的时候死讯了, 按照顺序是sasl crypto asn1 public_key ssl 发现是publi ...

  4. 如何开启多用户同时远程连接(Windows2008 Windows2012)

  5. openstack neutron网络主机节点网口配置 liberty版本之前的

  6. rhel及相关linux系统版本知识

    Rhel 此处Rhel非等同redhat哦,RedHat是红帽公司在1994年左右开发维护的linux桌面版本,在2004年左右红帽公司放弃redhat开始进军linux服务器版本开发,具体见下截图 ...

  7. 关于python requests包新版本设置代理的问题

    在更新了requests包之后,发现我电脑上的charles工具无法再成功抓取到数据包.百度了半年都没有找到原因. 然后 我使用了 google 查到了 charles的最新的文档发现.需要设置代理, ...

  8. [转]iSCSI完全指南

    [转]iSCSI完全指南 Posted on 2008-04-01 18:57 Tony Zhang 阅读(2102) 评论(0) 编辑 收藏 引:在上世纪末.本世纪初,一提到SAN(Storage ...

  9. 第二百零五天 how can I 坚持

    身体无论什么时候都是最重要的.肝血管瘤,头一次听说,应该没什么大碍.父母年龄大了,我们也该尽快把自己的事情处理好,让他们放心了. 规律作息,合理饮食,多注意锻炼. 该怎么办.不能这拖下去. 好好规划下 ...

  10. Java之对象池

    单例模式是限制了一个类只能有一个实例,对象池模式则是限制一个类实例的个数.对象池类就像是一个对象管理员,它以Static列表(也就是装对象的池子)的形式存存储某个实例数受限的类的实例,每一个实例还要加 ...