description: You have an array a consisting of n distinct positive integers, numbered from 1 to n. Define p_k as $$$p_k = ∑_{1 ≤ i, j ≤ k} a_i mod a_j, where x \bmod y denotes the remainder when x is divided by y. You have to find and print p_1, p_2, \ldots, p_n$$$. 
  
  Input
  
  The first line contains n — the length of the array (2 ≤ n ≤ 2 ⋅ 10^5).
  
  The second line contains n space-separated distinct integers a_1, …, a_n (1 ≤ a_i ≤ 3 ⋅ 10^5, a_i ≠ a_j if i ≠ j). 
  
  Output
  
  Print n integers p_1, p_2, …, p_n. 
  
  Examples
  
  Input
  
  
  4
  6 2 7 3
  
  
  Output
  
  
  0 2 12 22
  
  
  Input
  
  
  3
  3 2 1
  
  
  Output
  
  
  0 3 5
solution: #include <bits/stdc++.h>
  using namespace std;
  const int MX = 3E5;
  struct tree {
    long long T[MX * 4];
    void update(int t, int tl, int tr, int pos, int val) {
      if (tl > pos || tr < pos) return;
      if (tl == tr) {
        T[t] += val;
        return;
      }
      int tm = (tl + tr) / 2;
      update(t * 2 + 1, tl, tm, pos, val);
      update(t * 2 + 2, tm + 1, tr, pos, val);
      T[t] = T[t * 2 + 1] + T[t * 2 + 2];
    }
    long long query(int t, int tl, int tr, int l, int r) {
      if (tl > r || l > tr) return 0;
      if (l <= tl && tr <= r) return T[t];
      int tm = (tl + tr) / 2;
      return query(t * 2 + 1, tl, tm, l, r) + query(t * 2 + 2, tm + 1, tr, l, r);
    }
    long long get(int l, int r) { return query(0, 1, MX, l, r); }
    void upd(int pos, long long val) { update(0, 1, MX, pos, val); }
  };
  tree prefix_dif, cnt, sum;
  void solve() {
    int n;
    cin >> n;
    long long res = 0;
    for (int i = 1; i <= n; ++i) {
      int cur;
      cin >> cur;
      res += (i - 1 - cnt.get(1, cur)) * cur;
      res += cnt.get(1, cur) * cur - prefix_dif.get(1, cur);
      res += sum.get(1, cur);
      for (int mult = 1; mult * cur <= MX; ++mult) {
        long long num = mult * cur;
        long long nxt = min(MX, (mult + 1) * cur - 1);
        res += sum.get(num, nxt) - cnt.get(num, nxt) * num;
        prefix_dif.upd(num, cur);
      }
      cnt.upd(cur, 1);
      sum.upd(cur, cur);
      cout << res << ' ';
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t;
    t = 1;
    while (t--) solve();
    return 0;
  }
  