A. Alex and a Rhombus time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output While playing with geometric figures Alex has accidentally invented a concept of a nn-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≥2n≥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).
Input The first and only input line contains integer nn (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 nn-th order rhombus.
Examples inputCopy 1 outputCopy 1 inputCopy 2 outputCopy 5 inputCopy 3 outputCopy 13 Note Images of rhombus corresponding to the examples are given in the statement.
思路：找规律：a1=1，a2=1 4，a3=1 4 8，a4=1 4 8 12
故答案为an=1
=2*n*n-2*n 1
代码语言：javascript复制#include<bits/stdc .h>
#define rg register long long
#define inf 21474899993647
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
#define ll long long
#define maxn 100005
#define endl "n"
#define N 6000
const double eps=1e-8;
using namespace std;
inline ll read()
{
char ch=getchar();
ll s=0,w=1;
while(ch<48||ch>57)
{
if(ch=='-')
w=-1;
ch=getchar();
}
while(ch>=48&&ch<=57)
{
s=(s<<1) (s<<3) (ch^48);
ch=getchar();
}
return s*w;
}
inline void write(ll x)
{
if(x<0)
putchar('-'),x=-x;
if(x>9)
write(x/10);
putchar(x 48);
}
ll n=read();
ll ans=1;
int main()
{
ll tot=4;
while(n!=1)
{
ans =tot;
tot =4;
n--;
}
cout<<ans<<endl;
return 0;
}
