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 = 3e5 + 5;
  struct fenwick {
    long long c[N];
    void add(int i, int v) {
      for (; i < N; i += i & -i) {
        c[i] += v;
      }
    }
    long long qry(int i) {
      long long re = 0;
      for (; i > 0; i -= i & -i) {
        re += c[i];
      }
      return re;
    }
  } fw[2];
  int a[N];
  int main() {
    int n;
    scanf("%d", &n);
    long long ans = 0, sum = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      ans += sum;
      ans += 1ll * (i - 1) * a[i];
      for (int j = a[i]; j < N; j += a[i]) {
        int cnt = fw[0].qry(min(N - 1, j + a[i] - 1)) - fw[0].qry(j - 1);
        ans -= 1ll * cnt * j;
      }
      fw[0].add(a[i], 1);
      ans -= fw[1].qry(a[i]);
      for (int j = a[i]; j < N; j += a[i]) {
        fw[1].add(j, j);
        fw[1].add(j + a[i], -j);
      }
      sum += a[i];
      printf("%lld ", ans);
    }
    return 0;
  }
  