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 int maxn = 2e5 + 3;
  struct node {
    int a, b;
  } q[maxn], t[maxn];
  int cmp1(node x, node y) {
    if (x.a == y.a) return x.b < y.b;
    return x.a < y.a;
  }
  int n, tr[maxn], tot;
  void add(int x, int y) {
    for (int i = x; i < 2e5 + 3; i += i & (-i)) {
      tr[i] = max(tr[i], y);
    }
  }
  int que(int x) {
    int res = 0;
    for (int i = x; i; i -= i & (-i)) {
      res = max(res, tr[i]);
    }
    return res;
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &q[i].a);
      q[i].b = i - q[i].a;
    }
    sort(q + 1, q + n + 1, cmp1);
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      if (q[i].a != q[i - 1].a) {
        for (int j = 1; j <= tot; ++j) add(t[j].a, t[j].b);
        tot = 0;
      }
      if (q[i].b < 0) continue;
      int res = que(q[i].b + 1);
      t[++tot].a = q[i].b + 1;
      t[tot].b = res + 1;
      ans = max(ans, res + 1);
    }
    cout << ans << endl;
    return 0;
  }
  