Codeforces Round #313 (Div. 1) A. Gerald's Hexagon 数学题
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
- 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 ...
- 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 ...
- 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 ...
- 【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 tes ...
- Codeforces Round #313 (Div. 2) C. Gerald'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 ...
- Codeforces Round #313 (Div. 2) 560C Gerald's Hexagon(脑洞)
C. Gerald's Hexagon time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- 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 ...
- 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 ...
- 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
- 存储过程中使用事务与try catch
一.存储过程中使用事务的简单语法 在存储过程中使用事务时非常重要的,使用数据可以保持数据的关联完整性,在Sql server存储过程中使用事务也很简单,用一个例子来说明它的语法格式: 代码 : ) ) ...
- OSX学习02之更新输入法
OSX下最好的输入法是什么?话说在使用Windows的时候就知道了——它叫鼠须管. 想必大家用Windows的时候,进入系统第一步就是先装输入法吧~ OSX也是一样,自带输入法虽然凑合,但是我们作为A ...
- 《DevOps故障排除:Linux服务器运维最佳实践》读书笔记
首先,这本书是Linux.CN赠送的,多谢啦~ http://linux.cn/thread-12733-1-1.html http://linux.cn/thread-12754-1-1.html ...
- winform form
WinForm:Windows Form,.Net中用来开发Windows窗口程序的技术,无论是之前学的控制台程序,还是后面要学的asp.net都是调用.net框架,因此所有知识点都是一样的.新建一个 ...
- Chapter13:拷贝控制
拷贝控制操作:拷贝构造函数.拷贝赋值运算符.移动构造函数.移动赋值运算符.析构函数. 实现拷贝控制操作的最困难的地方是首先认识到什么时候需要定义这些操作. 拷贝构造函数: 如果一个构造函数的第一个参数 ...
- DouNet学习_Excel导入导出
Excel --->列是有限的 -->数据靠在单元格右边是数字类型,左边是字符串类型 把一个数字当初字符串来显示 在前面加个 ' -->用程序操作Excel 可以使用Excel的所有 ...
- stl 中List vector deque区别
stl提供了三个最基本的容器:vector,list,deque. vector和built-in数组类似,它拥有一段连续的内存空间,并且起始地址不变,因此 它能非常好的支持随 ...
- 单源最短路径-Dijkstra算法
1.算法标签 贪心 2.算法描述 具体的算法描述网上有好多,我觉得莫过于直接wiki,只说明一些我之前比较迷惑的. 对于Dijkstra算法,最重要的是维护以下几个数据结构: 顶点集合S : 表示已经 ...
- BFS寻路算法的实现
关于BFS的相关知识由于水平有限就不多说了,感兴趣的可以自己去wiki或者其他地方查阅资料. 这里大概说一下BFS寻路的思路,或者个人对BFS的理解: 大家知道Astar的一个显著特点是带有启发函数, ...
- geeksforgeeks@ Largest Number formed from an Array
http://www.practice.geeksforgeeks.org/problem-page.php?pid=380 Largest Number formed from an Array G ...