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 char nl = '\n';
  const long long inf = 1e9 + 10;
  const long long inf2 = 1e18 + 99LL;
  const long double inf3 = 1e17;
  const long long mod = 1e9 + 7, mod2 = 998244353;
  const long double eps = 1e-9;
  const bool local = false;
  const int logn = 17, maxn = 300001, maxm = 200001, maxn2 = 3;
  long long t[maxn * 2], t2[maxn * 2];
  void update(int i) {
    for (t[i += maxn]++; i > 1; i >>= 1) t[i >> 1] = t[i] + t[i ^ 1];
  }
  long long query(long long l, long long r) {
    long long res = 0;
    for (l += maxn, r += maxn; l < r; l >>= 1, r >>= 1) {
      if (l & 1) res += t[l++];
      if (r & 1) res += t[--r];
    }
    return res;
  }
  void update2(int l, int r, long long x) {
    for (l += maxn, r += maxn; l < r; l >>= 1, r >>= 1) {
      if (l & 1) t2[l++] += x;
      if (r & 1) t2[--r] += x;
    }
  }
  long long query2(int i) {
    long long res = 0;
    for (i += maxn; i; i >>= 1) res += t2[i];
    return res;
  }
  long long ans = 0;
  long long p[maxn], a[maxn];
  void f1(int i) {
    int l = a[i];
    ans += p[i];
    while (l < maxn) {
      ans -= query(l, min((long long)maxn, l + a[i])) * l;
      l += a[i];
    }
    update(a[i]);
  }
  void f2(int i) {
    ans += a[i] * i - query2(a[i]);
    int l = a[i];
    while (l < maxn) {
      update2(l, min((long long)maxn, l + a[i]), l);
      l += a[i];
    }
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    ;
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) p[i + 1] = p[i] + a[i];
    for (int i = 0; i < n; i++) {
      f1(i);
      f2(i);
      cout << ans << ' ';
    }
    cout << nl;
  }
  