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;
  const int maxn = 3e5 + 5;
  const int maxm = 1e5 + 5;
  const int inf = 0x3f3f3f3f;
  int read() {
    int s = 0, w = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) {
      if (c == '-') w = -1;
    }
    for (; isdigit(c); c = getchar()) {
      s = (s << 3) + (s << 1) + (c ^ 48);
    }
    return (w == 1 ? s : -s);
  }
  struct fenwick {
    long long tree[maxn];
    void add(int pos, long long k) {
      for (; pos < maxn; pos += (pos) & (-pos)) tree[pos] += k;
    }
    long long query(int pos) {
      long long res = 0;
      for (; pos; pos -= (pos) & (-pos)) res += tree[pos];
      return res;
    }
    long long query(int l, int r) { return query(r) - query(l - 1); }
  } f[2];
  int n;
  long long a[maxn];
  int main() {
    n = read();
    long long sum = 0, ans = 0;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      ans += sum;
      long long temp = 0, cnt = 0;
      for (int j = a[i]; j < maxn; j += a[i]) {
        cnt++;
        temp += cnt * f[0].query(j, min((long long)maxn - 1, j + a[i] - 1));
      }
      ans -= temp * a[i];
      f[0].add(a[i], 1);
      temp = (i - 1) * a[i];
      temp -= f[1].query(a[i]);
      ans += temp;
      for (int j = a[i]; j < maxn; j += a[i]) {
        f[1].add(j, j);
        f[1].add(min((long long)maxn - 1, j + a[i]), -j);
      }
      sum += a[i];
      printf("%lld\n", ans);
    }
    return 0;
  }
  