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;
  long long a[200005];
  struct BIT {
    long long tree[300005 << 1], Max = 6e5;
    void add(int x, long long d) {
      for (; x <= Max; x += x & -x) tree[x] += d;
    }
    long long ask(int x) {
      long long sum = 0;
      for (; x; x -= x & -x) sum += tree[x];
      return sum;
    }
    long long query(int l, int r) { return ask(r) - ask(l - 1); }
  } T1, T2;
  int main() {
    int n;
    scanf("%d", &n);
    long long sum = 0, ans = 0;
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    for (int i = 1; i <= n; i++) {
      sum += a[i];
      for (int j = a[i]; j <= 300005; j += a[i]) {
        ans += 1ll * j * T1.query(j, j + a[i] - 1);
        T2.add(j, a[i]);
      }
      ans += T2.ask(a[i]);
      T1.add(a[i], 1);
      printf("%lld ", 1ll * sum * i - ans);
    }
    return 0;
  }
  