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 long long maxx = 300005;
  long long n;
  long long a[200005];
  long long lowbit(long long x) { return x & (-x); }
  long long tree1[maxx], tree2[maxx];
  void add1(long long x, long long v) {
    while (x <= maxx) {
      tree1[x] += v;
      x += lowbit(x);
    }
  }
  void add2(long long x, long long v) {
    while (x <= maxx) {
      tree2[x] += v;
      x += lowbit(x);
    }
  }
  long long query1(long long x) {
    long long result = 0;
    while (x) {
      result += tree1[x];
      x -= lowbit(x);
    }
    return result;
  }
  long long query2(long long x) {
    long long result = 0;
    while (x) {
      result += tree2[x];
      x -= lowbit(x);
    }
    return result;
  }
  long long per, ans;
  signed main() {
    scanf("%lld", &n);
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      ans += per;
      per += a[i];
      for (long long j = a[i]; j <= maxx; j += a[i]) {
        ans -= (query1(min(j + a[i] - 1, maxx)) - query1(j - 1)) * j;
      }
      add1(a[i], 1);
      ans += (i - 1) * a[i] - query2(a[i]);
      for (long long j = a[i]; j <= maxx; j += a[i]) {
        add2(j, a[i]);
      }
      printf("%lld ", ans);
    }
    return 0;
  }
  