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 S = 800;
  const int MAXN = 200005;
  const int MAXM = 310000;
  int n;
  int a[MAXN];
  int cnt[S + 5];
  int cnt1[S + 5][S + 5];
  int cnt2[MAXM + 5];
  long long ans;
  long long mark1[MAXM + 5];
  void update(int &res, int l, int r, int v, bool flag = 0) {
    mark1[r] += v;
    if (flag) mark1[l - 1] -= v;
    res += cnt2[r] - (flag ? cnt2[l - 1] : 0);
  }
  int query(int l, int r, int v) {
    int res = 0;
    int L = l / S * S;
    if (L < l) L += S;
    if (L > r) {
      update(res, l, r, v, 1);
      return res;
    }
    update(res, l, L - 1, v, 1);
    while (L + S <= r) {
      update(res, L, L + S - 1, v);
      L += S;
    }
    update(res, L, r, v);
    return res;
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    long long TOT = 0, BIG = 0;
    for (int i = 1; i <= n; i++) {
      ans -= mark1[a[i]];
      for (int j = a[i] + 1; j % S != 0; j++) ans -= mark1[j];
      ans += (long long)BIG * a[i];
      if (a[i] <= S) {
        for (int j = 0; j < a[i]; j++) ans += (long long)j * cnt1[a[i]][j];
      } else {
        BIG++;
        ans += TOT;
        for (int j = 0; j * a[i] <= MAXM; j++)
          ans -= (long long)query(max(1, j * a[i]),
                                  min(MAXM, j * a[i] + a[i] - 1), j * a[i]) *
                 j * a[i];
      }
      for (int j = 1; j <= S; j++) ans += (long long)(a[i] % j) * cnt[j];
      if (a[i] <= S) cnt[a[i]]++;
      for (int j = 1; j <= S; j++) cnt1[j][a[i] % j]++;
      cnt2[a[i]]++;
      for (int j = a[i] + 1; j % S != 0; j++) cnt2[j]++;
      TOT += a[i];
      printf("%lld ", ans);
    }
    printf("\n");
    return 0;
  }
  