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 Max_N = 2e5, Inf = 1e9;
  int n, P[Max_N * 2 + 5];
  struct kk {
    int val, pos;
  } a[Max_N + 5];
  bool mycmp(kk x, kk y) {
    return x.val == y.val ? x.pos > y.pos : x.val < y.val;
  }
  int Ask(int x) {
    int ans = -Inf;
    for (x += n; x <= 2 * n; x += x & -x) ans = max(ans, P[x]);
    return ans;
  }
  void Add(int x, int val) {
    for (x += n; x; x -= x & -x) P[x] = max(P[x], val);
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i].val), a[i].pos = i;
    sort(a + 1, a + n + 1, mycmp);
    for (int i = 0; i <= 2 * n; i++) P[i] = -Inf;
    Add(0, 0);
    for (int i = 1; i <= n; i++) {
      int now = Ask(a[i].val - a[i].pos) + 1;
      Add(a[i].val - a[i].pos, now);
    }
    printf("%d\n", Ask(-n + 1));
    return 0;
  }
  