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 Fenwick {
    long long n;
    vector<long long> tree;
    Fenwick(long long N) : n(N + 1) { tree.resize(n); }
    void push(long long x, long long y) {
      for (long long i = x; i < n; i += (i & -i)) tree[i] += y;
    }
    long long get(long long x) {
      long long rv = 0;
      x = min(x, n - 1);
      for (long long i = x; i > 0; i -= (i & -i)) {
        rv += tree[i];
      }
      return rv;
    }
  };
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; ++i) {
      cin >> a[i];
    }
    const long long NS = (long long)3e5 + 4;
    Fenwick cnt(NS), sum(NS), lowsum(NS);
    long long ans = 0;
    for (long long i = 0; i < n; ++i) {
      long long lc = cnt.get(a[i] - 1);
      ans += lc * a[i] - lowsum.get(a[i]);
      ans += (cnt.get(NS) - cnt.get(a[i])) * a[i];
      ans += sum.get(NS);
      for (long long j = 1; j * a[i] < NS; ++j) {
        ans -= (cnt.get(a[i] * (j + 1) - 1) - cnt.get(a[i] * j - 1)) * a[i] * j;
        lowsum.push(a[i] * j, a[i]);
      }
      cnt.push(a[i], 1);
      sum.push(a[i], a[i]);
      cout << ans << ' ';
    }
    return 0;
  }
  