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 B = 600;
  int tcnt[300001];
  long long tree[300001], V[B + 1];
  void update(int n) {
    for (int v = n; n <= 300000; n++) {
      tree[n] += v;
      tcnt[n]++;
      if (n % B == B - 1) break;
    }
  }
  long long get_sum(int s, int e) {
    if (s / B == e / B) return tree[e] - (s % B ? tree[s - 1] : 0);
    long long ret = tree[s / B * B + B - 1] - (s % B ? tree[s - 1] : 0) + tree[e];
    s = s / B * B + B;
    e = e / B * B;
    while (s < e) {
      ret += tree[s + B - 1];
      s += B;
    }
    return ret;
  }
  int get_cnt(int s, int e) {
    if (s / B == e / B) return tcnt[e] - (s % B ? tcnt[s - 1] : 0);
    int ret = tcnt[s / B * B + B - 1] - (s % B ? tcnt[s - 1] : 0) + tcnt[e];
    s = s / B * B + B;
    e = e / B * B;
    while (s < e) {
      ret += tcnt[s + B - 1];
      s += B;
    }
    return ret;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ((void)0);
    ((void)0);
    ((void)0);
    int N, n;
    unsigned long long ans = 0;
    long long cv, cc;
    for (cin >> N; N--;) {
      int a;
      cin >> a;
      if (a <= B)
        ans += V[a];
      else {
        for (int i = 1; i <= a; i = n) {
          n = a / (a / i) + 1;
          cv = get_sum(i, n - 1);
          cc = get_cnt(i, n - 1);
          ans += cv;
          ans += 1LL * cc * a - (a / i) * cv;
        }
        for (int i = 2 * a;; i += a) {
          cv = get_sum(i - a, min(i - 1, 300000));
          cc = get_cnt(i - a, min(i - 1, 300000));
          ans += cv - 1LL * cc * (i - a);
          ans += 1LL * a * cc;
          if (i > 300000) break;
        }
      }
      for (int i = 1; i <= B; i++) V[i] += a % i + i % a;
      update(a);
      cout << ans << ' ';
    }
    return 0;
  }
  