数数

$dfsdfs$里面总共就两行，第一行很容易发现就是求$nn$，第二行就是每次迭代+2，总结就是让你求1，3，5，7……这样子的一个等差数列和，直接套公式就行。

#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
# include<bits/stdc++.h>
# include<unordered_map>

# define eps 1e-9
# define fi first
# define se second
# define ll long long
# define int ll
# define x1 sb
# define y1 dsb
# define x2 ssb
# define y2 ddsb
// cout<<fixed<<setprecision(n) 
//bool operator<(const Node& x )
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int > PII; 
const int mod=998244353;
const int N=5e5+10;
const int Time=86400;
const int X=131;
const int inf=0x3f3f3f3f;
const double PI = 1e-4;
double pai = 3.14159265358979323846; 
double e = exp(1);

int T,n,m,k,t,maxn,ans,p;


void solve(){
          cin >> n;
          ans = (1+2*n-1)*n/2;
		  cout<<ans; 
}	  
/*


*/
signed main(){  
    std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 
    
      //cin >> T;
    T = 1;
    while(T--){
		solve();
	} 
    return 0; 
}