Hexagons!（找规律）

Hexagons!
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Time Limit:500MS    
Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Status

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

题意：求不超过数字n的所有细胞总数。

当n=0时，有一个f（0）

当n=1时，外层是个6边形，可以看做每条边有2个细胞,不过每条边的第一个细胞和最后一个细胞被用2次。外层有：(!+1)*6-6

当n2时，外层也是个6变形，每条边有3个细胞,不过每条边的第一个细胞和最后一个细胞被用2次 .  外层有： (2+1)*6-6

""""""""""

n时，外层是个6变形，每条边有（n+1）个细胞,不过每条边的第一个细胞和最后一个细胞被用2次 .   外层共有：(n+1)*6-6个细胞

把上面每层细胞求和可得：

    sum=3*n*(n+1)+1;

My  solution:

/*2016.3.11*/

#include<stdio.h>
int main()
{
int i;
long long ans,n;
while(scanf("%I64d",&n)==1)
{
ans=3*n*(n+1)+1;
printf("%I64d\n",ans);
}
return 0;
}