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;
  void print(bool n) {
    if (n)
      cout << "YES";
    else
      cout << "NO";
    cout << '\n';
  }
  long long n;
  vector<long long> a;
  vector<long long> seg;
  void build(long long id = 1, long long l = 0, long long r = n) {
    if (r - l == 1) {
      seg[id] = a[l];
      return;
    }
    long long mid = (l + r) / 2;
    build(2 * id, l, mid);
    build(2 * id + 1, mid, r);
    seg[id] = max(seg[2 * id], seg[2 * id + 1]);
  }
  long long query(long long x, long long y, long long id = 1, long long l = 0,
                  long long r = n) {
    if (l >= x && r <= y) {
      return seg[id];
    }
    if (l >= y || r <= x) {
      return 0;
    }
    long long mid = (l + r) / 2;
    return max(query(x, y, 2 * id, l, mid), query(x, y, 2 * id + 1, mid, r));
  }
  void modify(long long pos, long long val, long long id = 1, long long l = 0,
              long long r = n) {
    if (r - l == 1) {
      seg[id] = val;
      a[pos] = val;
      return;
    }
    long long mid = (l + r) / 2;
    if (pos < mid)
      modify(pos, val, 2 * id, l, mid);
    else
      modify(pos, val, 2 * id + 1, mid, r);
    seg[id] = max(seg[2 * id], seg[2 * id + 1]);
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n;
    a.resize(n);
    seg.resize(4 * n + 1);
    long long b[n];
    for (long long i = 0; i < n; i++) {
      cin >> b[i];
      --b[i];
    }
    vector<vector<long long>> mp(n);
    for (long long i = 0; i < n; i++)
      if (i - b[i] >= 0 && b[i] < n) mp[b[i]].push_back(i - b[i]);
    for (const vector<long long> &x : mp)
      for (auto it = x.rbegin(); it != x.rend(); ++it)
        modify(*it, query(0, *it + 1) + 1);
    cout << query(0, n);
    return 0;
  }
  