Editorial

We solved D with Mo's Algorithm in under 2s. The idea is to have purely $$$O(Q \sqrt N)$$$ complexity (no log factor). You can do that with linked lists and undo (change inserions to erase and adapt Mo's Algo accordingly).

#include <bits/stdc++.h>

using namespace std;
const int BS = 128;

struct List {
  int n;
  vector<int> ops;
  vector<int> val2pos;
  vector<int> nxt, prv;
  long long ans = 0;

  List(const vector<int>& p) : n(p.size()), val2pos(n, -1), nxt(n, -1), prv(n, -1) {
    for (int i = 0; i < n; ++i) {
      val2pos[p[i]] = i; 
    }
    for (int i = 0; i < n; ++i) {
      if (i > 0) prv[i] = i - 1;
      if (i + 1 < n) nxt[i] = i + 1;
      ans += get(i, nxt[i]);
    }
  }

  int get(int a, int b) {
    if (a == -1 || b == -1) return 0;
    return abs(val2pos[a] - val2pos[b]);
  }

  long long Get() { 
    return ans; 
  }

  void Erase(int x) {
    ops.push_back(x);
    int nx = nxt[x], pr = prv[x];
    if (pr != -1) nxt[pr] = nx;
    if (nx != -1) prv[nx] = pr;
    ans += get(pr, nx) - get(pr, x) - get(x, nx);
  }
  void Undo() {
    int x = ops.back(); ops.pop_back();
    int nx = nxt[x], pr = prv[x];
    if (pr != -1) nxt[pr] = x;
    if (nx != -1) prv[nx] = x;
    ans -= get(pr, nx) - get(pr, x) - get(x, nx);
  }
  
};

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);


  int n, q; cin >> n >> q;
  vector<int> p(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i]; --p[i];
  }

  vector<vector<tuple<int, int, int>>> qrs(n);
  vector<long long> ans(q);

  for (int i = 0; i < q; ++i) {
    int l, r; cin >> l >> r; --l; --r;
    qrs[l / BS].emplace_back(r, l, i);
  }

  for (int i = 0; i < n; i += BS) {
    auto& qs = qrs[i / BS];
    sort(qs.rbegin(), qs.rend());
    List L(p);
    for (int j = 0; j < i; ++j)
      L.Erase(p[j]);

    int now_r = n - 1, now_l = i;
    for (auto [r, l, j] : qs) {
      while (now_r > r) L.Erase(p[now_r--]);
      while (now_l < l) L.Erase(p[now_l++]);
      ans[j] = L.Get();
      while (now_l > i) L.Undo(), now_l--;
    }
  }

  for (int i = 0; i < q; ++i) {
    cout << ans[i] << '\n';
  }

  return 0;
}

We used Mo algorithm with linked list over sorted segment. Deletion from linked list is $$$\mathcal{O}(1)$$$ and you use rollbacks to insert elements. To make this work with Mo algorithm sort queries by $$$(r/B, l, r)$$$. Then for each block you have $$$l \in [0, r)$$$ and $$$r \in [B * block, (B+1)*block)$$$, so insert all items from range $$$[0, (B+1)*block)$$$ into linked list, then to handle each query in sorted order you:

1. Permanently remove some elements from left side
2. Temporarily remove some elements from right side
3. Retrieve answer
4. Rollback all deletions from right side

My Solution for Div.2 M: Let's shuffle the given permutation. The interactor is not adaptive, so we can assume that each location has an equal probability of containing a treasure, assuming it is $$$p$$$. Suppose we examine the first 20 locations in turn, then the probability that the first 20 positions do not contain any treasure is $$$(1-p)^{20}$$$. If this happens, then we can assume that $$$p$$$ is sufficiently small, so that among the remaining positions, we can assume that the probability of picking any two positions, both of which contain the treasure is small ($$$p^2$$$). Then, we perform about 20 more queries, choosing 2 adjacent positions for each queries. There is a $$$p^2$$$ probability of missing the answer. But since $$$p$$$ is small enough, $$$p^2$$$ is also small enough. If we still do not find an answer, then we consider the distribution of the treasure to be more sparse, and continue to increase the number of locations per query. The limit on the number of queries does not appear to be strict, and can be passed by taking a random reasonable set of strategies. Code

Problem A~K are the same as Div.1 problems

Div.2 L: The Euler characteristic shows that $$$V-E+F-C=1$$$ on any planar graph $$$G$$$. So we have $$$C=V-E$$$ and the answer is simply $$$\max(n-m,0)$$$

Div.2 M: here

Div.2 N: After convert each portkey to intervals, the problem is essentially the same as CF786B. There will be up to $$$5n+m$$$ endpoints of the interval, but only $$$n+m$$$ of them will be useful. Code

Div.2 O: The distance of the two points is just $$$d=|x_1-y_1|+|x_2-y_2|$$$, and note that the given graph is a bipartite graph, so the given condition is equivalent to $$$d \geq w$$$ and $$$d\equiv w \pmod 2$$$.

The problem is that leaf $$$\lfloor m/2 \rfloor$$$ can be in the middle of a subtree. If you first scan half of the subtree, then later the other half, the bad guy could move between the two parts of the subtree.

Assume that the drunkard is at position $$$x$$$, and let $$$\mathrm{dp}_{i,j}$$$ denote the probability of being at moment $$$i$$$ and passing through position $$$x+j$$$ at least once. Obviously $$$x$$$ does not affect the answer, so we only need to focus on the relative distance we are from the end point at each moment. We have $$$dp_{i,0}=1$$$, and $$$\mathrm{dp}_{i,j}=p \cdot \mathrm{dp}_{i+1,j}+(1-p) \cdot \mathrm{dp}_{i+1,j-a_{i+1}}$$$, so we can calculate all $$$\mathrm{dp}_{*,*}$$$ in $$$O(n^2)$$$.

It's easy to observe that if the relative distance(signed) between the end point and the start point is $$$x$$$, then the probability of passing through $$$x$$$ at least once is just $$$\mathrm{dp}_{0,x}$$$, and the answer is $$$\frac{1}{2n+1} \sum_{i} \mathrm{dp}_{0,i}$$$. Code

You should probably ask one of closest sector's admin or register new sector. All information is here.

My teammates just checked 50 points around)) I don't know why it works.

Maybe someone from Div.1 will be able to explain this problem ?

google convex hull trick