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;
  using vi = vector<int>;
  using vvi = vector<vi>;
  using ii = pair<int, int>;
  using vii = vector<ii>;
  using vvii = vector<vii>;
  const int MOD = 1000000007;
  const int MX = 100001;
  int main() {
    int n;
    cin >> n;
    vii v;
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      if (i >= x) v.push_back(ii(i + 1 - x, x));
    }
    sort(v.begin(), v.end());
    vi s;
    for (auto i : v) {
      auto it = lower_bound(s.begin(), s.end(), i.second);
      if (it == s.end())
        s.push_back(i.second);
      else
        *it = i.second;
    }
    cout << (int)s.size();
    return 0;
  }
  