output_description: Print the number of ways to choose a pair of an even number and an odd number
  from the positive integers between 1 and K (inclusive).
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      K
attempt: #include <bits/stdc++.h>
  using namespace std;
  
  #define mem(arr, i) fill(arr, arr + sizeof(arr), i)
  #define MEM(arr, i) memset(arr, i, sizeof(arr))
  #define Push push_back
  #define Pair make_pair
  #define ALL(x) x.begin(), x.end()
  #define X first
  #define Y second
  #define ACCEL ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
  typedef unsigned long long ull;
  typedef long long lli;
  typedef pair<int,int> pii;
  typedef pair<lli,lli> pll;
  typedef pair<pll,lli> p3l;
  typedef vector<int> vcI;
  typedef vector<lli> vcL;
  typedef vector<pii> vPI;
  typedef vector<pll> vPL;
  typedef vector<p3l> v3L;
  typedef vector<vcL> vvL;
  //typedef priority_queue<T> prior_S<T>;
  //typedef priority_queue< T, vector<T>, greater<T> > prior_B<T>;
  const double PI = 3.141592653589793;
  const lli I_MAX = 1LL << 60;
        lli   mod = 1000000007;
  
  /// templates ///
  
  const lli maxN = 2E5;
  
  vcL inv(maxN+5, 1);
  vcL fac(maxN+5, 1);
  
  void ADD(lli &x, lli y) {
  	x += y;
  	if (x >= mod) x -= mod;
  	if (x < 0) x += mod;
  }
  
  void MUL(lli &x, lli y) {
  	x *= y;
  	x = (x % mod + mod) % mod;
  }
  
  lli mypow(lli b, lli e) {
  	lli ans = 1;
  	while (e) {
  		if (e & 1) MUL(ans, b);
  		MUL(b, b);
  		e >>= 1;
  	}
  	return ans;
  }
  
  lli modinv(lli n) {
  	return mypow(n, mod - 2);
  }
  
  void calcInv(lli n) {
  	for (int i = 2; i <= n; ++i) {
  		inv[i] = (mod - mod / i) * inv[mod % i] % mod;
  	}
  }
  
  void calcFac(lli n) {
  	for (int i = 2; i <= n; ++i) {
  		fac[i] = fac[i - 1] * i % mod;
  	}
  }
  
  lli cm(lli a, lli b) {
  	lli ans = 1;
  	if (a < b) return 0;
  
  	MUL(ans, inv[fac[a - b]]);
  	MUL(ans, inv[fac[b]]);
  	MUL(ans, fac[a]);
  
  	return ans;
  }
  
  lli Lucas(lli n, lli m) {
  	if (m == 0) return 1;
  	return cm(n % mod, m % mod) * Lucas(n / mod, m / mod) % mod;
  }
  
  lli doLucas(lli n, lli m) {
  	calcInv(mod);
  	calcFac(mod);
  	return Lucas(n, m);
  }
  
  /// coding area ///
  
  const lli maxn = 1E5;
  
  void precal() {
  
  	return;
  }
  
  void solve() {
  	lli n;
  	cin >> n;
  	cout << (n / 2) * (n - n / 2) << '\n';
  }
  
  int main() {
  	precal();
  	int t = 1;
  	#ifdef LOCAL
  		t = INT_MAX;
  	#else
  		ACCEL;
  	#endif
  //	cin >> t;
  	for (int i = 1; i <= t; ++i) {
  //		printf("Case #%d: ", i );
  		solve();
  	}
  	return 0;
  }
  
  
problem_description: Statement
  
  Find the number of ways to choose a pair of an even number and an odd number
  from the positive integers between 1 and K (inclusive). The order does not
  matter.
samples: [{"input": "3", "output": "2\n    \n\nTwo pairs can be chosen: (2,1) and (2,3).\n\n* * *"}, {"input": "6", "output": "9\n    \n\n* * *"}, {"input": "11", "output": "30\n    \n\n* * *"}, {"input": "50", "output": "625"}]