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;
  const long long dim = 200009, inf = 1e9;
  struct AINT {
    long long aint[4 * dim];
    void update(long long nod, long long tl, long long tr, long long pos,
                long long val) {
      if (tl == tr) {
        aint[nod] = max(aint[nod], val);
        return;
      }
      long long tm = (tl + tr) / 2;
      if (pos <= tm)
        update(2 * nod, tl, tm, pos, val);
      else
        update(2 * nod + 1, tm + 1, tr, pos, val);
      aint[nod] = max(aint[2 * nod], aint[2 * nod + 1]);
    }
    long long query(long long nod, long long tl, long long tr, long long l,
                    long long r) {
      if (tl == l && tr == r) {
        return aint[nod];
      }
      long long tm = (tl + tr) / 2;
      if (r <= tm) return query(2 * nod, tl, tm, l, r);
      if (l > tm) return query(2 * nod + 1, tm + 1, tr, l, r);
      return max(query(2 * nod, tl, tm, l, tm),
                 query(2 * nod + 1, tm + 1, tr, tm + 1, r));
    }
  } aint;
  struct elem {
    long long val, diff;
    bool operator<(const elem x) const {
      if (val == x.val) return diff > x.diff;
      return val < x.val;
    }
  };
  vector<elem> p;
  long long n, v[dim], ans;
  signed main() {
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      cin >> v[i];
      if (v[i] <= i) {
        p.push_back({v[i], i - v[i] + 1});
      }
    }
    sort(p.begin(), p.end());
    for (auto it : p) {
      long long nr = it.diff;
      long long val = aint.query(1, 1, n, 1, nr);
      ans = max(ans, val + 1);
      aint.update(1, 1, n, nr, val + 1);
    }
    cout << ans;
  }
  