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;
  using ll = long long;
  using ld = long double;
  template <class T>
  bool uin(T& a, const T& b) {
    return a > b ? a = b, 1 : 0;
  }
  template <class T>
  bool uax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const int N = 3e5 + 10;
  template <class S, class T>
  pair<S, T> operator+(const pair<S, T>& a, const pair<S, T>& b) {
    return {a.first + b.first, a.second + b.second};
  }
  struct {
    using node = pair<ll, ll>;
    node t[N << 1];
    void activate(int p) {
      t[p + N] = {p, 1};
      for (p += N; p >>= 1;) t[p] = t[p << 1] + t[p << 1 | 1];
    }
    node query(int l, int r) {
      r = min(r, N);
      node res = {0, 0};
      for (l += N, r += N; l < r; l >>= 1, r >>= 1) {
        if (l & 1) res = res + t[l++];
        if (r & 1) res = res + t[--r];
      }
      return res;
    }
  } one;
  struct {
    ll t[N << 1];
    ll query(int p) {
      ll res = 0;
      for (p += N; p > 0; p >>= 1) res += t[p];
      return res;
    }
    void update(int l, int r, int del) {
      r = min(r, N);
      for (l += N, r += N; l < r; l >>= 1, r >>= 1) {
        if (l & 1) t[l++] += del;
        if (r & 1) t[--r] += del;
      }
    }
  } two;
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    vector<int> a(n);
    for (auto& x : a) cin >> x;
    vector<ll> ans(n);
    for (auto k = (0); k < (n); ++k) {
      if (k) ans[k] = ans[k - 1];
      ans[k] += 1ll * k * a[k] - two.query(a[k]);
      for (int l = 0; l < N; l += a[k]) {
        auto [sum, cnt] = one.query(l, l + a[k]);
        ans[k] += sum - cnt * l;
        two.update(l, l + a[k], l);
      }
      one.activate(a[k]);
    }
    for (auto x : ans) cout << x << ' ';
    cout << '\n';
  }
  