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 ull = long unsigned long;
  using ld = double long;
  template <class T, int N, int B>
  struct asym_ra {
    T a[N + 1];
    T b[N / B + 2];
    T get(int r) {
      if (r <= 0) return T();
      if (r > N) r = N;
      return a[r] + b[r / B];
    }
    T operator()(int l, int r) { return get(r) - get(l); }
    void add(int p, const T& v) {
      for (int i = p + 1; i % B && i <= N; i++) a[i] += v;
      for (int i = p / B + 1; i < N / B + 2; i++) b[i] += v;
    }
  };
  int n;
  int a[200005];
  const int H = 300000;
  asym_ra<ll, H + 5, 512> cnt;
  asym_ra<ll, H + 5, 512> sum;
  int main() {
    ios::sync_with_stdio(!cin.tie(0));
    cin >> n;
    ll sol = 0;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      for (int j = 0; j <= H; j += x) {
        sol += sum(j, j + x) - cnt(j, j + x) * 1ll * j;
      }
      for (int q = 1;;) {
        int r = x / q;
        int l = x / (q + 1) + 1;
        r++;
        sol += x * cnt(l, r) - q * sum(l, r);
        if (l == 1) break;
        q = x / (l - 1);
      }
      sol += x * 1ll * cnt(x + 1, H + 1);
      cout << sol << ' ';
      cnt.add(x, 1);
      sum.add(x, x);
    }
    cout << '\n';
  }
  