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 i, j, k, n, m, t, f[200500], sb;
  void add(int x, int y) {
    for (; x <= 200000; x += (-x & x)) {
      f[x] = max(f[x], y);
    }
  }
  int get(int x, int y = 0) {
    for (; x; x -= (-x & x)) {
      y = max(y, f[x]);
    }
    return y;
  }
  set<int> s[400500];
  int main() {
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
      scanf("%d", &k);
      if (k > i) continue;
      s[i + 200000 - k].insert(k);
    }
    for (i = 1; i <= 400000; i++) {
      for (auto j : s[i]) {
        sb = get(j - 1) + 1;
        add(j, sb);
      }
    }
    cout << get(200000);
  }
  