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 N = 300000;
  int n;
  long long ans = 0;
  long long T1[N + 5];
  long long T2[N + 5];
  long long T3[N + 5];
  void upd(long long *T, int x, int v) {
    while (x <= N) {
      T[x] += v, x += x & (-x);
    }
  }
  long long qry(long long *T, int x) {
    long long ret = 0;
    while (x > 0) ret += T[x], x -= x & (-x);
    return ret;
  }
  long long query(long long *T, int l, int r) {
    r = min(r, N);
    return qry(T, r) - qry(T, l - 1);
  }
  void update(int num, int th) {
    for (int i = 0; i * num <= N; ++i) {
      ans += query(T1, i * num, (i + 1) * num - 1) -
             query(T2, i * num, (i + 1) * num - 1) * i * num;
    }
    ans += 1ll * num * (th - 1) - query(T3, 1, num);
    upd(T1, num, num);
    upd(T2, num, 1);
    for (int i = num; i <= N; i += num) {
      upd(T3, i, num);
    }
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      int num;
      scanf("%d", &num);
      update(num, i);
      printf("%lld ", ans);
    }
    return 0;
  }
  