D -
Hexagons!

Crawling in process...
Crawling failed
Time Limit:500MS    
Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Description

After a probationary period in the game development company of IT City Petya was included in a group of the programmers that develops a new turn-based strategy game resembling the well known "Heroes of Might & Magic". A part of the game is turn-based fights
of big squadrons of enemies on infinite fields where every cell is in form of a hexagon.

Some of magic effects are able to affect several field cells at once, cells that are situated not farther than
n cells away from the cell in which the effect was applied. The distance between cells is the minimum number of cell border crosses on a path from one cell to another.

It is easy to see that the number of cells affected by a magic effect grows rapidly when
n increases, so it can adversely affect the game performance. That's why Petya decided to write a program that can, given
n, determine the number of cells that should be repainted after effect application, so that game designers can balance scale of the effects and the game performance. Help him to do it. Find the number of hexagons situated
not farther than n cells away from a given cell.

Input

The only line of the input contains one integer n (0 ≤ n ≤ 109).

Output

Output one integer — the number of hexagons situated not farther than
n cells away from a given cell.

Sample Input

Input
2
Output
19
1,6,12,18....明显就是1,1*6,2*6,3*6....除了第一项,就是一个等差序列带公式就行了
#include<iostream>
using namespace std;
#define LL long long
int main()
{
LL n;
cin>>n;
cout<<n*(6+6*n)/2+1<<endl;
return 0;
}


Codeforces--630D--Hexagons(规律)的更多相关文章

  1. codeforces 630D Hexagons!

    D. Hexagons! time limit per test 0.5 seconds memory limit per test 64 megabytes input standard input ...

  2. codeforces 615E Hexagons (二分+找规律)

    E. Hexagons time limit per test 1 second memory limit per test 256 megabytes input standard input ou ...

  3. codeforces 362A找规律

    刚开始以为是搜索白忙活了原来是个简单的找规律,以后要多想啊 此题是两马同时跳 A. Two Semiknights Meet time limit per test 1 second memory l ...

  4. codeforces——思路与规律

    codeforces 804B     http://codeforces.com/problemset/problem/804/B /* 题意:给定一个只含ab的序列,每次操作可将ab变为bba 问 ...

  5. Nastya and a Wardrobe CodeForces - 992C(规律)

    写一下二叉树  推一下公式就出来了, 注意取模时的输出形式 #include <bits/stdc++.h> #define mem(a, b) memset(a, b, sizeof(a ...

  6. Codeforces 145A-Lucky Conversion(规律)

    A. Lucky Conversion time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  7. codeforces D. Queue 找规律+递推

    题目链接: http://codeforces.com/problemset/problem/353/D?mobile=true H. Queue time limit per test 1 seco ...

  8. Codeforces Gym 100015A Another Rock-Paper-Scissors Problem 找规律

    Another Rock-Paper-Scissors Problem 题目连接: http://codeforces.com/gym/100015/attachments Description S ...

  9. Codeforces Round #338 (Div. 2) E. Hexagons 讨论讨论

    E. Hexagons 题目连接: http://codeforces.com/contest/615/problem/E Description Ayrat is looking for the p ...

  10. Codeforces Round #347 (Div. 2) C. International Olympiad 找规律

    题目链接: http://codeforces.com/contest/664/problem/C 题解: 这题最关键的规律在于一位的有1989-1998(9-8),两位的有1999-2098(99- ...

随机推荐

  1. 高级Java知识

    高级Java知识(JVM.字节码.内存模型) 内存=方法区+栈空间+堆+程序计数器 栈(stack)包括虚拟机栈(VM stack)和本地方法栈(native method stack). 方法区(m ...

  2. 初识关系型数据库(SQL)与非关系型数据库(NOSQL)

    一.关系型数据库(SQL): Mysql,oracle 特点:数据和数据之间,表和字段之间,表和表之间是存在关系的 例如:部门表 001部分,   员工表 001 用户表,用户名.密码 分类表 和 商 ...

  3. 使用doxmate生成文档

    主页:http://html5ify.com/doxmate/ 在windows下面使用doxmate 1. 下载node.js(msi)并安装 http://www.nodejs.org/downl ...

  4. SSL协议提供的服务

    SSL协议提供的服务主要有: 1)认证用户和服务器,确保数据发送到正确的客户机和服务器: 2)加密数据以防止数据中途被窃取: 3)维护数据的完整性,确保数据在传输过程中不被改变.

  5. css知识框架

  6. Self-Attetion

    四.self-attention 1.是什么? attention机制通常用在encode与decode之间,但是self-attention则是输入序列与输出序列相同,寻找序列内部元素的关系即 K= ...

  7. 阅读《JavaScript设计模式》第三章心得

    简单工厂模式 1.通过类实例化对象创建 传统的用面向对象方法去创建很多类去实现某些功能不妥当,这样不仅占用的很多类名称,而且别人使用这些方法的同时要记住每个类的名字,所以这样不适合团队开发,所以我们可 ...

  8. [luogu2154 SDOI2009] 虔诚的墓主人(树状数组+组合数)

    传送门 Solution 显然每个点的权值可以由当前点上下左右的树的数量用组合数\(O(1)\)求出,但这样枚举会T 那么我们考虑一段连续区间,对于一行中两个常青树中间的部分左右树的数量一定,我们可用 ...

  9. Maven中更改默认JDK版本

    只要在settings.xml文件中加上如下标签即可.(我这里是默认的1.7版本) <profiles> <profile> <id>jdk-1.7</id& ...

  10. Django-cookie与session操作

    添加cookie: def login(req): if req.method=="POST": uf = UserInfoForm(req.POST) if uf.is_vali ...