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 = 300000 + 50;
  int n;
  long long a[200000 + 50];
  long long P[200000 + 50];
  struct Seg1 {
    int l, r;
    long long lzy;
    long long sum;
  } s[2][N << 2];
  struct BIT {
    long long s[N], n;
    void init(int m) {
      n = m;
      memset(s, 0, (n + 1) << 2);
    }
    void add(int p, long long d) {
      while (p <= n) {
        s[p] += d;
        p += (p & -p);
      }
    }
    long long query(int p) {
      long long ret = 0;
      while (p > 0) {
        ret += s[p];
        p -= (p & -p);
      }
      return ret;
    }
  } bit[2];
  inline void build(int id, int p, int l, int r) {
    s[id][p].l = l, s[id][p].r = r;
    if (l == r) {
      s[id][p].sum = s[id][p].lzy = 0;
      return;
    }
    int mid = (l + r) >> 1;
    build(id, p << 1, l, mid);
    build(id, p << 1 | 1, mid + 1, r);
  }
  inline void pushdown(int id, int p) {
    if (s[id][p].lzy != 0) {
      s[id][p << 1].lzy += s[id][p].lzy;
      s[id][p << 1 | 1].lzy += s[id][p].lzy;
      s[id][p << 1].sum += (s[id][p << 1].r - s[id][p << 1].l + 1) * s[id][p].lzy;
      s[id][p << 1 | 1].sum +=
          (s[id][p << 1 | 1].r - s[id][p << 1 | 1].l + 1) * s[id][p].lzy;
      s[id][p].lzy = 0;
    }
  }
  inline void add(int id, int p, int l, int r, long long d) {
    int L = s[id][p].l, R = s[id][p].r;
    if (L == l && R == r) {
      s[id][p].lzy += d;
      s[id][p].sum += (R - L + 1) * d;
      return;
    }
    pushdown(id, p);
    int mid = (L + R) >> 1;
    if (r <= mid)
      add(id, p << 1, l, r, d);
    else if (l > mid)
      add(id, p << 1 | 1, l, r, d);
    else
      add(id, p << 1, l, mid, d), add(id, p << 1 | 1, mid + 1, r, d);
    s[id][p].sum = s[id][p << 1].sum + s[id][p << 1 | 1].sum;
  }
  inline void change(int id, int p, int pos, long long d) {
    int L = s[id][p].l, R = s[id][p].r;
    if (L == R) {
      s[id][p].sum += d;
      return;
    }
    int mid = (L + R) >> 1;
    if (pos <= mid)
      change(id, p << 1, pos, d);
    else
      change(id, p << 1 | 1, pos, d);
    s[id][p].sum = s[id][p << 1].sum + s[id][p << 1 | 1].sum;
  }
  inline long long query(int id, int p, int l, int r) {
    int L = s[id][p].l, R = s[id][p].r;
    if (L == l && R == r) return s[id][p].sum;
    pushdown(id, p);
    int mid = (L + R) >> 1;
    if (r <= mid)
      return query(id, p << 1, l, r);
    else if (l > mid)
      return query(id, p << 1 | 1, l, r);
    else
      return query(id, p << 1, l, mid) + query(id, p << 1 | 1, mid + 1, r);
  }
  int main() {
    build(0, 1, 1, 300000);
    bit[0].init(300000);
    bit[1].init(300000);
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%lld", &a[i]);
      if (a[i] > 1) P[i] += bit[0].query(a[i] - 1);
      if (a[i] < 300000)
        P[i] += a[i] * (bit[1].query(300000) - bit[1].query(a[i]));
      if (a[i] > 1)
        P[i] += a[i] * bit[1].query(a[i] - 1) - query(0, 1, a[i], a[i]);
      if (a[i] < 300000) {
        for (int j = 1;; ++j) {
          int l = (j == 1) ? a[i] + 1 : a[i] * j;
          int r = min(300000ll, (j + 1) * a[i] - 1);
          if (r < l) continue;
          P[i] += bit[0].query(r) - bit[0].query(l - 1) -
                  (bit[1].query(r) - bit[1].query(l - 1)) * j * a[i];
          add(0, 1, l, r, j * a[i]);
          if (r == 300000) break;
        }
      }
      P[i] += P[i - 1];
      bit[0].add(a[i], a[i]);
      bit[1].add(a[i], 1);
      printf("%lld ", P[i]);
    }
    return 0;
  }
  