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 N = 3e5 + 7, S = sqrt(N);
  int mx = 0, Bs, B;
  struct BLOCK {
    long long sum[N], val[N];
    inline void add(int x, int y) {
      int sy = (x + B - 1) / B;
      for (int i = (sy); i <= (Bs); ++i) sum[i] += y;
      for (int i = (x); i <= (min(sy * B, mx)); ++i) val[i] += y;
    }
    inline long long qzh(int x) { return x ? sum[(x - 1) / B] + val[x] : 0; }
    inline long long query(int l, int r) { return qzh(r) - qzh(l - 1); }
  } bl1, bl2;
  int n, a[N], p;
  long long f[N];
  inline long long calc1(int k) {
    long long res = 0;
    for (int l = 1, r; l <= k; l = r + 1)
      r = k / (k / l), res += (k / l) * bl2.query(l, r);
    return res;
  }
  inline long long calc2(int k) {
    long long res = 0;
    if (k <= B) return f[k];
    for (int i = (1); i <= (mx / k); ++i)
      res += i * bl1.query(i * k, min((i + 1) * k - 1, mx));
    return res;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long ns = 0, sum = 0;
    cin >> n;
    for (int i = (1); i <= (n); ++i) cin >> a[i], mx = max(mx, a[i]);
    B = sqrt(mx), Bs = (mx + B - 1) / B;
    for (int i = (1); i <= (n); ++i) {
      p = i, ns += (long long)a[i] * (i - 1) + sum;
      ns -= calc1(a[i]);
      ns -= calc2(a[i]) * a[i];
      cout << ns << " ";
      sum += a[i];
      bl1.add(a[i], 1), bl2.add(a[i], a[i]);
      for (int j = (1); j <= (B); ++j) f[j] += a[i] / j;
    }
    return 0;
  }
  