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;
  int a[500010];
  pair<int, int> b[500010];
  int c[500010];
  int T, n, tot;
  int main(void) {
    T = 1;
    while (T--) {
      scanf("%d", &n);
      for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
        b[i] = {a[i], -i};
      }
      sort(b + 1, b + 1 + n);
      for (int i = 1; i <= n; ++i) {
        if (-b[i].second >= b[i].first) {
          int x = -(b[i].second + b[i].first);
          int pos = upper_bound(c + 1, c + 1 + tot, x) - c;
          c[pos] = x;
          tot = max(pos, tot);
        }
      }
      printf("%d\n", tot);
    }
    return 0;
  }
  