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>
  #pragma GCC optimize("O3")
  #pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
  using namespace std;
  template <typename val1>
  void __f(const char* name, val1&& value) {
    cout << name << " = " << value << endl;
  }
  template <typename val1, typename... values>
  void __f(const char* names, val1&& value, values&&... multiplevalue) {
    const char* comma = strchr(names + 1, ',');
    cout.write(names, comma - names) << " = " << value << " ";
    __f(comma + 1, multiplevalue...);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n;
    vector<int> trees(2 * n, 0);
    int arr[n + 1];
    function<int(int)> query = [&](int a) {
      int ans = 0;
      while (a) {
        ans = max(trees[a], ans);
        a = a - (a & (-a));
      }
      return ans;
    };
    function<void(int, int)> update = [&](int a, int v) {
      while (a < 2 * n) {
        trees[a] = max(trees[a], v);
        a = a + (a & (-a));
      }
    };
    int ans = 0;
    for (int i = 1; i <= n; i++) cin >> arr[i];
    vector<int> index(n);
    iota(index.begin(), index.end(), 1);
    sort(index.begin(), index.end(), [&](int a, int b) {
      if (arr[a] == arr[b]) return a > b;
      return arr[a] < arr[b];
    });
    for (int i = 0; i < n; i++) {
      if (index[i] < arr[index[i]]) continue;
      int off = index[i] - arr[index[i]] + 1;
      int fc = query(off);
      ans = max(ans, fc + 1);
      update(off, fc + 1);
    }
    cout << ans << endl;
    cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
         << "\n";
    return 0;
  }
  