思路：单调栈（分别统计左边小于等于当前大小的数量）

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef double db;
typedef long double ldb;
typedef pair<int, int> pii;
typedef pair<ll, ll> PII;
#define pb emplace_back
//#define int ll
#define all(a) a.begin(),a.end()
#define x first
#define y second
#define ps push_back
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)

void solve();

const int N = 1e6 + 10;


signed main() {
    IOS;
    solve();
    return 0;
}

void solve() {
    ll n; cin >> n;
    vector<ll> a(n+1),ans1(n+1,0),ans2(n+1,0);
    stack<ll> st;
    for(int i = 1; i <= n; ++ i) cin >> a[i];
    for(int i = 1; i <= n; ++ i)
    {
        while(st.size() && st.top() > a[i]) st.pop();
        ans1[i] = st.size();
        st.push(a[i]);
    }
    while(st.size()) st.pop();
    for(int i = n; i >= 1; -- i)
    {
        while(st.size() && st.top() > a[i]) st.pop();
        ans2[i] = st.size();
        st.push(a[i]);
    }
    for(int i = 1; i <= n; ++ i) cout << ans1[i] + ans2[i] << " \n"[i==n];
}