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;
  using ll = long long;
  template <class T = int>
  T rd() {
    T res = 0;
    T fg = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') fg = -1;
      ch = getchar();
    }
    while (isdigit(ch)) res = (res << 1) + (res << 3) + (ch ^ 48), ch = getchar();
    return res * fg;
  }
  const int N = 300010;
  template <typename T = int>
  struct Fenwick {
    const int n;
    T t[N];
    Fenwick(int n) : n(n) { memset(t, 0, sizeof t); }
    void add(int k, T v) {
      for (; k <= n; k += k & -k) t[k] += v;
    }
    T qsum(int k) {
      T v = 0;
      for (; k; k -= k & -k) v += t[k];
      return v;
    }
    T get(int l, int r) { return qsum(r) - qsum(l - 1); }
  };
  int n;
  int a[N];
  int main() {
    n = rd();
    for (int i = 1; i <= n; i++) a[i] = rd();
    Fenwick<ll> f1(300000), f2(300000);
    ll pre = 0, ans = 0;
    for (int i = 1; i <= n; i++) {
      ans += 1ll * a[i] * (i - 1);
      ans += pre;
      ans -= f1.qsum(a[i]);
      for (int j = a[i]; j <= 300000; j += a[i]) {
        int l = j, r = min(300000, j + a[i] - 1);
        ans -= f2.get(l, r) * j;
        f1.add(l, a[i]);
      }
      f2.add(a[i], 1);
      pre += a[i];
      printf("%lld%c", ans, " \n"[i == n]);
    }
    return 0;
  }
  