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;
  long long n, a1, a2, sa, M = 3e5, ak, s1[300005], s2[300005];
  void f1(long long p, long long q) {
    while (p <= M) {
      s1[p] += q;
      p += (p & -p);
    }
  }
  void f2(long long p, long long q) {
    while (p <= M) {
      s2[p] += q;
      p += (p & -p);
    }
  }
  long long g1(long long p) {
    long long re = 0;
    while (p >= 1) {
      re += s1[p];
      p -= (p & -p);
    }
    return re;
  }
  long long g2(long long p) {
    long long re = 0;
    while (p >= 1) {
      re += s2[p];
      p -= (p & -p);
    }
    return re;
  }
  int main() {
    long long i, j, k, t, st, la;
    cin >> n;
    for (k = 1; k <= n; k++) {
      scanf("%lld", &ak);
      a1 += (k - 1) * ak;
      a1 -= g1(ak);
      a2 += sa;
      st = 0;
      for (i = 0;; i++) {
        la = min((i + 1) * ak - 1, M);
        t = g2(la);
        a2 -= (t - st) * ak * i;
        st = t;
        if ((i + 1) * ak - 1 >= M) break;
      }
      sa += ak;
      for (i = 1; i * ak <= M; i++) {
        f1(i * ak, ak);
      }
      f2(ak, 1);
      printf("%lld ", a1 + a2);
    }
    return 0;
  }
  