Mega Inversions
Para cada , calculamos la cantidad de y de forma independiente.
Con indexed sets
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
template <class T>
using Tree =
tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i = a; i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i, a, b) for (int i = (b) - 1; i >= a; i--)
#define F0Rd(i, a) for (int i = (a) - 1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
const int MOD = 1000000007;
int n;
ll hi[100000], lo[100000];
vi z;
ll ans = 0;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
z.resize(n);
F0R(i, n) cin >> z[i];
Tree<pii> T1;
F0R(i, n) {
hi[i] = T1.size() - T1.order_of_key({z[i], MOD});
T1.insert({z[i], i});
}
Tree<pii> T2;
F0Rd(i, n) {
lo[i] = T2.order_of_key({z[i], -MOD});
T2.insert({z[i], i});
}
F0R(i, n) ans += lo[i] * hi[i];
cout << ans;
}Con Árboles de Fenwick (BIT)
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MX = 2e5 + 5;
ll bit2[MX], bit1[MX];
int n;
void upd(int i, int val, ll ar[]) {
for (; i <= n; i += i & (-i)) { ar[i] += val; }
}
ll query(int i, ll ar[]) {
ll res = 0;
for (; i; i -= i & (-i)) { res += ar[i]; }
return res;
}
int main() {
cin >> n;
vector<int> ar(n);
for (int i = 0; i < n; i++) { cin >> ar[i]; }
ll sol = 0;
for (int i = n - 1; i >= 0; i--) {
sol += query(ar[i] - 1, bit2);
ll uno = query(ar[i] - 1, bit1);
upd(ar[i], 1, bit1);
upd(ar[i], uno, bit2);
}
cout << sol << '\n';
}