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>
  #pragma GCC optimize(2)
  using namespace std;
  const int inf = 1000000000;
  long long mod = 1000000007LL;
  long long mod2 = 998244353LL;
  int t, n, m;
  long long bit[2][300005];
  int a[300005];
  int lowbit(int x) { return x & (-x); }
  void update(int id, int pos, long long val) {
    while (pos < 300005) {
      bit[id][pos] += val;
      pos += lowbit(pos);
    }
  }
  long long query(int id, int pos) {
    long long res = 0LL;
    while (pos) {
      res += bit[id][pos];
      pos -= lowbit(pos);
    }
    return res;
  }
  int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
    }
    long long sum = 0LL;
    long long tot = 0LL;
    for (int i = 1; i <= n; ++i) {
      long long res = (long long)(i - 1) * (long long)a[i] - query(1, a[i]);
      res += sum;
      for (int j = 1; j <= 300000 / a[i]; j++) {
        long long det =
            query(0, min(300000, (j + 1) * a[i] - 1)) - query(0, j * a[i] - 1);
        res -= (long long)a[i] * (long long)det * (long long)j;
      }
      tot += res;
      printf("%lld", tot);
      if (i < n)
        cout << " ";
      else
        cout << endl;
      sum += a[i];
      update(0, a[i], 1);
      for (int j = 1; j <= 300000 / a[i]; ++j) {
        update(1, a[i] * j, (long long)j * (long long)a[i]);
        update(1, min(300000 + 1, a[i] * (j + 1)),
               -(long long)j * (long long)a[i]);
      }
    }
    return 0;
  }
  