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;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    const int M = 3e5;
    int n;
    cin >> n;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    vector<int64_t> fen1(2 * M + 1), fen2(2 * M + 1);
    int64_t pref = 0, s = 0, t = 0;
    auto Sum = [](int i, vector<int64_t>& fen) -> int64_t {
      int64_t r = 0;
      while (i > 0) {
        r += fen[i];
        i -= i & (-i);
      }
      return r;
    };
    auto Upd = [](int i, vector<int64_t>& fen, int d) -> void {
      while (i < (int)fen.size()) {
        fen[i] += d;
        i += i & (-i);
      }
    };
    for (int i = 1; i <= n; i++) {
      s += 1LL * a[i] * (i - 1) - Sum(a[i], fen1);
      t += pref;
      pref += a[i];
      for (int j = 1; j * a[i] <= 300000; j++) {
        t -= 1LL * (Sum((j + 1) * a[i] - 1, fen2) - Sum(j * a[i] - 1, fen2)) * j *
             a[i];
        Upd(a[i] * j, fen1, j * a[i]);
        Upd(a[i] * (j + 1), fen1, -j * a[i]);
      }
      Upd(a[i], fen2, 1);
      cout << s + t << " ";
    }
    return 0;
  }
  