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;
  struct FenwickTree {
    vector<long long int> bit;
    int n;
    FenwickTree(int n) {
      this->n = n;
      bit.assign(n, 0);
    }
    FenwickTree(vector<long long int> a) : FenwickTree(a.size()) {
      for (size_t i = 0; i < a.size(); i++) add(i, a[i]);
    }
    long long int sum(int r) {
      long long int ret = 0;
      for (; r >= 0; r = (r & (r + 1)) - 1) ret += bit[r];
      return ret;
    }
    long long int sum(int l, int r) { return sum(r) - sum(l - 1); }
    void add(int idx, long long int delta) {
      for (; idx < n; idx = idx | (idx + 1)) bit[idx] += delta;
    }
  };
  const int MAX_VAL = 3e5;
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    vector<long long int> a(n);
    vector<long long int> ra(MAX_VAL + 5, -1);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      ra[a[i]] = i;
    }
    vector<long long int> tr(n);
    vector<long long int> ctr(MAX_VAL + 5);
    int cidx = 0;
    for (int i = 1; i <= MAX_VAL; ++i) {
      if (ra[i] != -1) {
        tr[ra[i]] = cidx;
        cidx++;
      }
      ctr[i] = cidx - 1;
    }
    long long int csum = 0;
    long long int tsum = 0;
    FenwickTree fw(n);
    FenwickTree fw2(n);
    for (int i = 0; i < n; ++i) {
      csum += tsum + a[i] * fw2.sum(tr[i]);
      for (long long int k = 1; k * k <= a[i]; ++k) {
        csum -= fw.sum(ctr[a[i] / k]);
        if (a[i] / k > k) {
          csum -= (a[i] / k - a[i] / (k + 1)) * fw.sum(ctr[k]);
        }
      }
      for (long long int k = 2; k * a[i] <= MAX_VAL; ++k) {
        csum -= a[i] * (i - fw2.sum(ctr[k * a[i] - 1]));
      }
      cout << csum << " ";
      tsum += a[i];
      fw.add(tr[i], a[i]);
      fw2.add(tr[i], 1);
    }
    cout << endl;
  }
  