#include <iostream>
using namespace std;
#define endl '\n'

const int N = 1e5 + 5;
int n, dp[N] = {1, 1, 2};
signed main()
{
  cin >> n;
  for (int i = 3; i <= n; i++)
    dp[i] = dp[i - 3] + i / 2 + 1;
  cout << dp[n];
  return 0;
}