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;
  template <typename T>
  struct BIT {
    int n;
    vector<T> bit;
    BIT(int _n) : n(_n), bit(_n + 1) {}
    void add(int x, T val) {
      for (; x <= n; x += x & -x) bit[x] += val;
    }
    T get(int x) {
      T ret = 0;
      for (; x > 0; x -= x & -x) ret += bit[x];
      return ret;
    }
    T query(int l, int r) {
      if (l > r) {
        return 0;
      }
      return get(r) - get(l - 1);
    }
  };
  const int BLOCK = 550;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    int m = *max_element(a.begin(), a.end());
    BIT<long long> bit_sum(2 * m);
    BIT<int> bit_cnt(2 * m);
    vector<int> cnt_block(BLOCK);
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      int x = a[i];
      for (int j = x, k = 0; k <= m; j += x, k += x) {
        ans +=
            (bit_sum.query(k + 1, j - 1) - 1LL * bit_cnt.query(k + 1, j - 1) * k);
      }
      ans += (1LL * bit_cnt.query(x + 1, m) * x);
      for (int j = 1; j * j <= x; j++) {
        int r = x / j;
        if (r < BLOCK) {
          break;
        }
        int l = max(BLOCK, x / (j + 1) + 1);
        if (l > r) {
          continue;
        }
        ans += (1LL * x * bit_cnt.query(l, r) - 1LL * j * bit_sum.query(l, r));
      }
      for (int j = 1; j < min(x, BLOCK); j++) {
        if (cnt_block[j] == 0) {
          continue;
        }
        ans += 1LL * (x % j) * cnt_block[j];
      }
      if (x < BLOCK) {
        cnt_block[x]++;
      }
      bit_sum.add(a[i], a[i]);
      bit_cnt.add(a[i], 1);
      cout << ans << " \n"[i == n - 1];
    }
    return 0;
  }
  