description: Mr. Chanek gives you a sequence a indexed from 1 to n. Define f(a) as the number of indices where a_i = i. 
  
  You can pick an element from the current sequence and remove it, then concatenate the remaining elements together. For example, if you remove the 3-rd element from the sequence [4, 2, 3, 1], the resulting sequence will be [4, 2, 1]. 
  
  You want to remove some elements from a in order to maximize f(a), using zero or more operations. Find the largest possible f(a).
  
  Input
  
  The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the initial length of the sequence.
  
  The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 2 ⋅ 10^5) — the initial sequence a.
  
  Output
  
  Output an integer denoting the largest f(a) that can be obtained by doing zero or more operations.
  
  Examples
  
  Input
  
  
  7
  2 1 4 2 5 3 7
  
  
  Output
  
  
  3
  
  
  Input
  
  
  4
  4 2 3 1
  
  
  Output
  
  
  2
  
  Note
  
  In the first example, f(A) = 3 by doing the following operations.
  
  [2,1,4,2,5,3,7] → [2,1,2,5,3,7] → [1,2,5,3,7] → [1,2,5,3] → [1,2,3]
  
  In the second example, f(A) = 2 and no additional operation is needed.
solution: #include <bits/stdc++.h>
  using namespace std;
  struct FenwickTree2D {
    vector<vector<int>> nodes;
    vector<vector<int>> f;
    int n;
    FenwickTree2D() {}
    FenwickTree2D(int n) : n(n) {
      nodes.resize(n + 1);
      f.resize(n + 1);
    }
    void fakeUpdate(int u, int v) {
      if (u == 0 || v == 0) {
        return;
      }
      for (int x = u; x <= n; x += x & -x) {
        nodes[x].push_back(v);
      }
    }
    void fakeGet(int u, int v) {
      if (u == 0 || v == 0) {
        return;
      }
      for (int x = u; x > 0; x -= x & -x) nodes[x].push_back(v);
    }
    void update(int u, int v, int delta) {
      if (u == 0 || v == 0) {
        return;
      }
      for (int x = u; x <= n; x += x & -x) {
        for (int y = lower_bound(nodes[x].begin(), nodes[x].end(), v) -
                     nodes[x].begin() + 1;
             y <= nodes[x].size(); y += y & -y) {
          f[x][y] = max(f[x][y], delta);
        }
      }
    }
    int get(int u, int v) {
      if (u == 0 || v == 0) {
        return 0;
      }
      int res = 0;
      for (int x = u; x > 0; x -= x & -x)
        for (int y = upper_bound(nodes[x].begin(), nodes[x].end(), v) -
                     nodes[x].begin();
             y > 0; y -= y & -y)
          res = max(res, f[x][y]);
      return res;
    }
    void build() {
      for (int i = 1; i <= n; ++i) {
        nodes[i].push_back(1000111000);
        sort(nodes[i].begin(), nodes[i].end());
        f[i].resize(nodes[i].size() + 3);
      }
    }
  };
  const int MAX = 2e5 + 5;
  int n, a[MAX], b[MAX], f[MAX];
  FenwickTree2D fw = FenwickTree2D(MAX);
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      b[i] = i - a[i] + 1;
      fw.fakeUpdate(a[i], b[i]);
      fw.fakeGet(a[i], b[i]);
    }
    fw.build();
    int rs = 0;
    for (int i = 1; i <= n; i++)
      if (b[i] > 0) {
        f[i] = fw.get(a[i] - 1, b[i]) + 1;
        rs = max(rs, f[i]);
        fw.update(a[i], b[i], f[i]);
      }
    cout << rs;
  }
  