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 long long Mod = 1000000007LL;
  const int N = 2e5 + 10;
  const long long Inf = 2242545357980376863LL;
  const long long Log = 30;
  int fen[N];
  void Add(int idx, int x) {
    idx++;
    for (; idx < N; idx += (idx & (-idx))) fen[idx] = max(fen[idx], x);
  }
  int Get(int idx) {
    int res = -N;
    idx++;
    for (; idx; idx -= (idx & (-idx))) res = max(fen[idx], res);
    return res;
  }
  vector<int> V[N];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    int v;
    for (int i = 1; i <= n; i++) {
      cin >> v;
      V[v].push_back(i);
    }
    fill(fen, fen + N, -N);
    Add(0, 0);
    int ans = 0;
    for (int i = 1; i < N; i++) {
      reverse(V[i].begin(), V[i].end());
      for (auto idx : V[i]) {
        if (idx < i) continue;
        int res = Get(idx - i) + 1;
        ans = max(ans, res);
        Add(idx - i, res);
      }
    }
    cout << ans << '\n';
    return 0;
  }
  