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 long long N = 300000;
  long long f[N + 2];
  void updf(long long x, long long d) {
    if (x > N) return;
    for (; x <= N; x += (x & -x)) f[x] += d;
  }
  long long cntf(long long x) {
    long long ans = 0;
    for (x = min(x, N); x > 0; x -= (x & -x)) ans += f[x];
    return ans;
  }
  long long fn[N + 2];
  void updfn(long long x, long long d) {
    if (x > N) return;
    for (; x <= N; x += (x & -x)) fn[x] += d;
  }
  long long cntfn(long long x) {
    long long ans = 0;
    for (x = min(x, N); x > 0; x -= (x & -x)) ans += fn[x];
    return ans;
  }
  long long g[N + 2];
  void updg(long long x, long long d) {
    if (x > N) return;
    for (; x <= N; x += (x & -x)) g[x] += d;
  }
  long long cntg(long long x) {
    long long ans = 0;
    for (x = min(x, N); x > 0; x -= (x & -x)) ans += g[x];
    return ans;
  }
  int main() {
    long long n;
    scanf("%lld", &n);
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      long long x;
      scanf("%lld", &x);
      ans += x * i - cntg(x);
      for (long long j = 0; j <= N; j += x)
        ans += cntf(j + x - 1) - cntf(j) - j * (cntfn(j + x - 1) - cntfn(j));
      printf("%lld ", ans);
      updfn(x, 1);
      updf(x, x);
      for (long long j = x; j <= N; j += x) {
        updg(j, j);
        updg(j + x, -j);
      }
    }
    printf("\n");
    return 0;
  }
  