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;
  using ld = long double;
  template <typename T>
  struct Fenwick {
    int n;
    vector<T> bit;
    Fenwick(int _n) : n(_n), bit(n + 1) {}
    void update(int p, T v) {
      for (++p; p <= n; p += p & -p) bit[p] += v;
    }
    T query(int p) {
      T ret{};
      for (++p; p > 0; p -= p & -p) ret += bit[p];
      return ret;
    }
  };
  int main() {
    int n;
    scanf("%d", &n);
    vector<int> A(n);
    const int MAX = 300000;
    for (int i = 0; i < n; ++i) {
      scanf("%d", &A[i]);
    }
    Fenwick<int> fcnt(MAX + 1);
    Fenwick<ll> rangeFenw(MAX + 1);
    ll curSum = 0, prefix = 0;
    vector<ll> ans(n);
    for (int i = 0; i < n; ++i) {
      curSum += (ll)A[i] * i - rangeFenw.query(A[i]);
      curSum += prefix;
      prefix += A[i];
      for (int j = 0; j <= MAX; j += A[i]) {
        int k = min(MAX, j + A[i] - 1);
        int cnt = fcnt.query(k) - fcnt.query(j - 1);
        curSum -= (ll)cnt * j;
        rangeFenw.update(j, j);
        rangeFenw.update(k + 1, -j);
      }
      fcnt.update(A[i], 1);
      42;
      ans[i] = curSum;
    }
    for (int i = 0; i < n; ++i) {
      printf("%lld ", ans[i]);
    }
  }
  