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 H {
    int i, x, y;
  };
  int k, m, n, b[200004];
  H w[200004];
  int C(H a, H b) { return a.x < b.x || a.x == b.x && a.i > b.i; }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    fill(b + 1, b + n + 1, 1 << 30);
    for (int i = 1; i <= n; i++) {
      cin >> k;
      w[i] = {i, k, i - k};
    }
    sort(w + 1, w + n + 1, C);
    for (int i = 1; i <= n; i++) {
      if (w[i].y >= 0) b[upper_bound(b + 1, b + n + 1, w[i].y) - b] = w[i].y;
    }
    for (int i = 1; i <= n; i++) {
      if (b[i] < 1 << 30) m = i;
    }
    cout << m;
  }
  