AtCode nadafes2022_day1 A - Max Inversion

N

3

2

#include <bits/stdc++.h>
#include <cstdlib>

using namespace std;

#define endl '\n';

int main() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    // 输入
    long long n;
    cin >> n;
    cout << fixed;
    if(n % 2 == 0) {
        cout << (n - 1) * n / 2;
    } else {
        cout << (n - 1) * n / 2 - 1;
    }

    return 0;
}

void solve() {
	ll n; 
	cin >> n;
	ll ans = n * (n - 1) / 2;
	if (n % 2) ans--;
	cout << ans << "\n";
}
 
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	solve();
	return 0;
}

using ll = long long;
#define endl '\n'
#define lfs cout<<fixed<<setprecision(10)
int main(){
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  ll res = 0,buf = 0;
  bool judge = true;
  ll n;
  cin >> n;
  cout << n * (n-1) / 2 - n % 2 << endl;
  return 0;
}