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 maxA = 3e5 + 5;
  long long sum[maxA], cnt[maxA];
  void updateSum(int pos, long long val) {
    pos++;
    while (pos < maxA) {
      sum[pos] += val;
      pos += pos & (-pos);
    }
  }
  long long querySum(int pos) {
    pos++;
    long long ans = 0;
    while (pos > 0) {
      ans += sum[pos];
      pos -= pos & (-pos);
    }
    return ans;
  }
  void updateCnt(int pos, long long val) {
    pos++;
    while (pos < maxA) {
      cnt[pos] += val;
      pos += pos & (-pos);
    }
  }
  long long queryCnt(int pos) {
    pos++;
    long long ans = 0;
    while (pos > 0) {
      ans += cnt[pos];
      pos -= pos & (-pos);
    }
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    fill(sum, sum + maxA, 0);
    fill(cnt, cnt + maxA, 0);
    long long a[n];
    for (int x = 0; x < n; x++) {
      cin >> a[x];
    }
    long long ans = 0, sum = 0;
    for (int x = 0; x < n; x++) {
      ans += sum + a[x] * x + querySum(a[x]);
      for (int y = a[x]; y < maxA; y += a[x]) {
        updateSum(y, -a[x]);
        ans -= (queryCnt(maxA - 3) - queryCnt(y - 1)) * a[x];
      }
      updateCnt(a[x], 1);
      sum += a[x];
      if (x > 0) cout << " ";
      cout << ans;
    }
    cout << "\n";
    return 0;
  }
  