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 int N = 2e5 + 5;
  int n, tr[N];
  int lowbit(int x) { return x & (-x); }
  void update(int x, int t) {
    for (int i = x; i <= n; i += lowbit(i)) tr[i] = max(tr[i], t);
  }
  int ask(int x) {
    int ans = 0;
    for (int i = x; i > 0; i -= lowbit(i)) ans = max(ans, tr[i]);
    return ans;
  }
  set<int> e[N];
  int main() {
    scanf("%d", &n);
    for (int i = 1, x; i <= n; i++) {
      scanf("%d", &x);
      int j = i - x;
      if (j >= 0 && j <= i - 1) e[j].insert(x);
    }
    for (int i = 0; i <= n; i++) {
      for (auto x : e[i]) {
        int j = ask(x - 1);
        update(x, j + 1);
      }
    }
    printf("%d", ask(n));
    return 0;
  }
  