CodeForces-630 D. Hexagons!【递推】

最新推荐文章于 2021-03-24 21:16:19 发布

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最新推荐文章于 2021-03-24 21:16:19 发布

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分类专栏：简单数学知识 =====codeforces=====

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本文介绍了一个计算六边形网格中特定范围内单元格数量的方法。针对一款类似“英雄无敌”的策略游戏中，魔法效果作用范围内的六边形单元格数量计算问题，提供了一个高效的算法实现。

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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.

$\text{[math]}$

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.

Examples

input
2

output
19

题意：如图所示的六边形，求第n圈有多少个六边形

不难推导出公式

3*n*(n+1)+1

#include<stdio.h>
typedef long long ll;
int main()
{
    ll n;
    while(~scanf("%lld",&n))
    {
        printf("%lld\n",3*n*(n+1)+1);
    }
	return 0;
}