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 BIT {
    vector<long long> sum;
    BIT(int _n) : sum(_n + 2, 0) {}
    void update(int x, long long v) {
      for (++x; x < (int)sum.size(); x += x & -x) sum[x] += v;
    }
    long long get(int x) {
      long long ret = 0;
      for (++x; x; x -= x & -x) ret += sum[x];
      return ret;
    }
  };
  const int M = 3e5 + 5;
  void solve() {
    int n;
    scanf("%d", &n);
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
    }
    BIT sum(M), cnt(M);
    long long ans = 0, tot = 0;
    for (int i = 0; i < n; ++i) {
      int v = a[i];
      ans += tot + 1LL * v * i - sum.get(v);
      long long last = 0;
      for (int u = v; u < M + v; u += v) {
        long long cur = cnt.get(min(u - 1, M));
        ans -= (u - v) * (cur - last);
        last = cur;
      }
      cnt.update(v, +1);
      for (int u = v; u < M; u += v) {
        sum.update(u, v);
      }
      tot += v;
      printf("%lld%c", ans, i + 1 == n ? '\n' : ' ');
    }
  }
  int main() {
    int tc = 1;
    while (tc--) {
      solve();
    }
    return 0;
  }
  