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 FT {
    vector<long long> s;
    FT(int n) : s(n) {}
    void update(int pos, long long dif) {
      for (; pos < (int)(s).size(); pos |= pos + 1) s[pos] += dif;
    }
    long long query(int pos) {
      long long res = 0;
      for (; pos > 0; pos &= pos - 1) res += s[pos - 1];
      return res;
    }
    long long query(int l, int r) { return query(r) - query(l); }
  };
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < (n); ++i) cin >> a[i];
    int MAX_A = *(max_element(begin(a), end(a)));
    FT A(MAX_A + 1);
    FT B(MAX_A + 1);
    long long p = 0;
    long long pref_sum = 0;
    for (int i = 0; i < n; i++) {
      p += pref_sum;
      p += (long long)i * a[i];
      p -= A.query(a[i] + 1);
      for (int j = a[i]; j <= MAX_A; j += a[i]) {
        int l = j, r = min(MAX_A + 1, j + a[i]);
        p -= B.query(l, r) * j;
        A.update(l, a[i]);
      }
      B.update(a[i], 1);
      cout << p << " ";
      pref_sum += a[i];
    }
    return 0;
  }
  