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;
  mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
  mt19937 rnf(2106);
  const int N = 200005;
  int n;
  pair<int, int> a[N];
  int maxu[N * 10];
  void ubd(int tl, int tr, int x, int y, int pos) {
    if (tl == tr) {
      maxu[pos] = max(maxu[pos], y);
      return;
    }
    int m = (tl + tr) / 2;
    if (x <= m)
      ubd(tl, m, x, y, pos * 2);
    else
      ubd(m + 1, tr, x, y, pos * 2 + 1);
    maxu[pos] = max(maxu[pos * 2], maxu[pos * 2 + 1]);
  }
  int qry(int tl, int tr, int l, int r, int pos) {
    if (l > r) return -N;
    if (tl == l && tr == r) return maxu[pos];
    int m = (tl + tr) / 2;
    return max(qry(tl, m, l, min(m, r), pos * 2),
               qry(m + 1, tr, max(m + 1, l), r, pos * 2 + 1));
  }
  void solv() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i].first;
    for (int i = 1; i <= n; ++i) {
      a[i].second = i;
      a[i].first = (i - a[i].first);
    }
    sort(a + 1, a + n + 1);
    for (int i = 0; i < N * 10; ++i) maxu[i] = -N;
    ubd(0, N * 2, N + 0, 0, 1);
    for (int i = 1; i <= n; ++i) {
      if (a[i].first < 0) continue;
      ubd(0, N * 2, N + a[i].first - a[i].second,
          qry(0, N * 2, N + a[i].first - a[i].second + 1, N * 2, 1) + 1, 1);
    }
    cout << maxu[1] << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int tt = 1;
    while (tt--) solv();
    return 0;
  }
  