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 N = 3e5 + 5;
  struct FenwickTree {
    vector<long long> bit;
    int n;
    FenwickTree(int n) {
      this->n = n;
      bit.assign(n, 0);
    }
    long long query(int r) {
      long long ret = 0;
      for (; r >= 0; r = (r & (r + 1)) - 1) ret += bit[r];
      return ret;
    }
    long long query(int l, int r) { return query(r) - query(l - 1); }
    void rupd(int idx, long long delta) {
      for (; idx < n; idx = idx | (idx + 1)) bit[idx] += delta;
    }
  };
  void solve() {
    int n, x;
    long long res = 0;
    cin >> n;
    FenwickTree cnt(N), b(N), pre(N);
    for (int i = 0; i < n; i++) {
      cin >> x;
      res += 1ll * i * x - b.query(x);
      for (int j = 0; j < N; j += x) {
        int nxt = min(j + x - 1, N - 1);
        long long sum = pre.query(j, nxt), e = cnt.query(j, nxt);
        res += sum - e * j;
      }
      for (int j = x; j < N; j += x) b.rupd(j, x);
      cnt.rupd(x, 1);
      pre.rupd(x, x);
      cout << res << " ";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    solve();
    return 0;
  }
  