思路：单调栈（分别统计左边小于等于当前大小的数量）

#include   using namespace std; typedef long long ll; typedef double db; typedef long double ldb; typedef pair pii; typedef pair 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 a(n+1),ans1(n+1,0),ans2(n+1,0);     stack 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]; }