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 N = 3e5 + 7;
  const int sq = sqrt(300000);
  int n;
  int a[N], bl[N], pre1[N], Pre1[N];
  long long val[N], pre2[N], Pre2[N];
  void insert(int x) {
    for (int i = x; bl[i] == bl[x]; ++i) ++pre1[i], pre2[i] += x;
    for (int i = bl[x]; i <= bl[300000]; ++i) ++Pre1[i], Pre2[i] += x;
  }
  int ask1(int x) {
    if (x <= 0) return 0;
    return Pre1[bl[x] - 1] + pre1[x];
  }
  long long ask2(int x) {
    if (x <= 0) return 0;
    return Pre2[bl[x] - 1] + pre2[x];
  }
  int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = 1; i <= 300000; ++i) bl[i] = (i - 1) / sq + 1;
    long long ans = 0;
    for (int i = 1; i <= n; ++i) {
      for (int l = 1, r; l <= a[i]; l = r + 1) {
        r = a[i] / (a[i] / l);
        int k = a[i] / l;
        int num = ask1(r) - ask1(l - 1);
        long long sum = ask2(r) - ask2(l - 1);
        ans += 1ll * num * a[i] - sum * k;
      }
      ans += 1ll * (ask1(300000) - ask1(a[i])) * a[i];
      if (a[i] > sq) {
        for (int l = 0; l <= 300000; l += a[i]) {
          int r = min(l + a[i] - 1, 300000);
          int num = ask1(r) - ask1(l - 1);
          long long sum = ask2(r) - ask2(l - 1);
          ans += sum - 1ll * (l / a[i]) * a[i] * num;
        }
      } else
        ans += val[a[i]];
      for (int j = 1; j <= sq; ++j) val[j] += a[i] % j;
      insert(a[i]);
      printf("%lld ", ans);
    }
    return 0;
  }
  