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 MAXN = 3e5 + 7;
  struct nd {
    long long coeff, frq;
    long long sum;
  };
  nd tr[4 * MAXN];
  long long lazy[4 * MAXN];
  nd combine(const nd &a, const nd &b) {
    nd rt;
    rt.coeff = a.coeff + b.coeff;
    rt.frq = a.frq + b.frq;
    rt.sum = a.sum + b.sum;
    return rt;
  }
  void propagate(int u, int l, int r) {
    if (lazy[u]) {
      tr[u].sum += lazy[u] * tr[u].frq;
      tr[u].coeff += lazy[u];
      if (l != r) {
        lazy[u * 2] += lazy[u];
        lazy[u * 2 + 1] += lazy[u];
      }
      lazy[u] = 0;
    }
  }
  void update(int u, int l, int r, int x, int y, long long v) {
    propagate(u, l, r);
    if (r < x || y < l) return;
    if (x <= l && r <= y) {
      lazy[u] += v;
      propagate(u, l, r);
      return;
    }
    int mid = (l + r) / 2;
    update(u * 2, l, mid, x, y, v);
    update(u * 2 + 1, mid + 1, r, x, y, v);
    tr[u] = combine(tr[u * 2], tr[u * 2 + 1]);
  }
  void update(int u, int l, int r, int idx) {
    propagate(u, l, r);
    if (r < idx || idx < l) return;
    if (l == r) {
      tr[u].frq++;
      tr[u].sum += tr[u].coeff;
      return;
    }
    int mid = (l + r) / 2;
    update(u * 2, l, mid, idx);
    update(u * 2 + 1, mid + 1, r, idx);
    tr[u] = combine(tr[u * 2], tr[u * 2 + 1]);
  }
  nd query(int u, int l, int r, int x, int y) {
    propagate(u, l, r);
    if (x <= l && r <= y) return tr[u];
    int mid = (l + r) / 2;
    if (y <= mid) return query(u * 2, l, mid, x, y);
    if (mid < x) return query(u * 2 + 1, mid + 1, r, x, y);
    return combine(query(u * 2, l, mid, x, y),
                   query(u * 2 + 1, mid + 1, r, x, y));
  }
  long long a[MAXN];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    long long mx = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      mx = max(mx, a[i]);
    }
    long long presum = 0;
    for (int k = 1; k <= n; k++) {
      for (int x = a[k]; x <= mx; x += a[k]) {
        update(1, 1, mx, x, mx, a[k]);
      }
      update(1, 1, mx, a[k]);
      presum += a[k];
      cout << presum * k - tr[1].sum << " ";
    }
    return 0;
  }
  