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>
  const int N = 200010;
  int stk[N];
  std::pair<int, int> pr[N];
  int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(NULL);
    std::cout.tie(NULL);
    int n;
    std::cin >> n;
    for (int i = 1; i <= n; ++i) {
      int ai;
      std::cin >> ai;
      pr[i] = std::make_pair(ai, -i);
    }
    std::sort(pr + 1, pr + n + 1);
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      int x = -(pr[i].first + pr[i].second);
      if (x < 0) continue;
      int pos = std::upper_bound(stk + 1, stk + ans + 1, x) - stk;
      stk[pos] = x;
      ans = std::max(ans, pos);
    }
    std::cout << ans << '\n';
    return 0;
  }
  