Codeforces Round #569 (Div. 2)
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1×1 (i.e just a cell).
A n-th order rhombus for all n≥2 one obtains from a n−1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1≤n≤100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
input
1
output
1
input
2
output
5
input
3
output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
思路：感觉是找规律题，推一下找找规律就好了......
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<cmath>
5 #include<algorithm>
6 #include<map>
7 #include<set>
8 #include<vector>
9 #include<queue>
10 using namespace std;
11 #define ll long long
12 const int mod=1e9+7;
13 const int inf=1e9+7;
14
15 //const int maxn=
16
17 int main()
18 {
19 ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
20
21 int n;
22
23 while(cin>>n)
24 {
25 int ans=0;
26
27 int f=(n<<1)-1;
28
29 ans+=f;
30
31 f-=2;
32
33 while(f!=-1)
34 {
35 ans+=(f<<1);
36 f-=2;
37 }
38
39 cout<<ans<<endl;
40
41 }
42
43 return 0;
44 }