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;
  int read() {
    int a = 0;
    char c = getchar();
    while (!isdigit(c)) {
      c = getchar();
    }
    while (isdigit(c)) {
      a = a * 10 + c - 48;
      c = getchar();
    }
    return a;
  }
  const long long _ = 300003;
  long long N;
  struct BIT {
    long long sum[_];
    void add(long long x, long long v) {
      while (x <= 3e5) {
        sum[x] += v;
        x += ((x) & -(x));
      }
    }
    long long qry(long long x) {
      long long A = 0;
      while (x) {
        A += sum[x];
        x -= ((x) & -(x));
      }
      return A;
    }
    long long qry(long long l, long long r) { return qry(r) - qry(l - 1); }
  } sum1, sumn;
  signed main() {
    long long all = 0, ans = 0;
    N = read();
    for (long long i = 1; i <= N; ++i) {
      long long cnt = 0, A = read();
      ans += all + (i - 1) * A - sumn.qry(A);
      all += A;
      for (long long p = 1; p * A <= 3e5; ++p) {
        cnt += sum1.qry(p * A, 3e5);
        sumn.add(p * A, A);
      }
      ans -= cnt * A;
      printf("%lld ", ans);
      sum1.add(A, 1);
    }
    return 0;
  }
  