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>
  const int MAXN = 3e5 + 15;
  using namespace std;
  long long fwt[MAXN + 5][3];
  long long A[MAXN];
  void update(int k, int x, long long val) {
    while (x <= MAXN) {
      fwt[x][k] += val;
      x += (x & (-x));
    }
  }
  long long rsum(int k, int x) {
    long long su = 0;
    while (x >= 1) {
      su += fwt[x][k];
      x -= (x & (-x));
    }
    return su;
  }
  void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> A[i + 1];
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      long long x = A[i];
      if (x == 1) {
        update(0, x, 1);
        update(1, x, x);
      } else {
        for (int k = 0; k < MAXN; k += x) {
          int l = k + 1;
          int r = min(MAXN + 0LL, k + x - 1);
          ans += rsum(1, r) - rsum(1, l - 1) - (rsum(0, r) - rsum(0, l - 1)) * k;
        }
        update(0, x, 1);
        update(1, x, x);
      }
      ans += x * (i - 1) - (rsum(2, x));
      for (int k = 0; k * x < MAXN; k++) {
        int l = max(k * x, 1LL);
        int r = min(k * x + x - 1, 0LL + MAXN);
        update(2, r + 1, -k * x);
        update(2, l, k * x);
      }
      cout << ans << " ";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);
    cout << fixed << setprecision(12);
    int t = 1;
    for (int i = 1; i <= t; i++) {
      solve();
      cout << "\n";
    }
    return 0;
  }
  