当 $n$ 为偶数时，水平方向走的步数是 $k = \frac{n}{2}$，水平方向的位置有 $k + 1$ 种：

  • 左走 $0$ 步，右走 $k$ 步
  • 左走 $1$ 步，右走 $k - 1$ 步
  • …
  • 左走 $k$ 步，右走 $0$ 步

竖直方向同理。此时答案为 $k^2 = (\frac{n}{2})^2$。

当 $n = 2k + 1$ 为奇数时，分两种情况：

  1. 水平方向 $k$ 步，竖直方向 $k + 1$ 步。
  2. 水平方向 $k + 1$ 步，竖直方向 $k$ 步。

先考虑第 1 种情况，最终在水平方向的位置可能为：$k, k - 2, k - 4, \ldots$，可以发现和 $k$ 的奇偶性相同，所以两种情况是不重合的。第 1 种情况的答案是 $(k + 1)(k + 2)$，所以两种情况总共 $2(k + 1)(k + 2)$。

// Date: Mon Dec 18 21:31:45 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

template <typename t> istream &operator>>(istream &in, vector<t> &vec) {
  for (t &x : vec)
    in >> x;
  return in;
}

template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) {
  int n = SZ(vec);
  For(i, 0, n) {
    out << vec[i];
    if (i < n - 1)
      out << ' ';
  }
  return out;
}

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

int n;

int main(void) {
#ifdef _DEBUG
  freopen("1459b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  while (cin >> n) {
    int k = n / 2, res = 0;

    if (n & 1) {
      res = 2 * (k + 1) * (k + 2);
    } else {
      res = (k + 1) * (k + 1);
    }

    cout << res << '\n';
  }

  return 0;
}