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;
  inline int gi() {
    char tmp = getchar();
    int ans = 0, flag = 1;
    while (!isdigit(tmp)) {
      if (tmp == '-') flag = -1;
      tmp = getchar();
    }
    while (isdigit(tmp)) {
      ans = ans * 10 + tmp - '0';
      tmp = getchar();
    }
    return ans * flag;
  }
  inline void in(int &x) { x = gi(); }
  inline void in2(int &x, int &y) { in(x), in(y); }
  inline void in3(int &x, int &y, int &z) { in(x), in(y), in(z); }
  inline void in4(int &x, int &y, int &z, int &a) { in3(x, y, z), in(a); }
  inline void smax(int &x, int y) { x = max(x, y); }
  inline void smin(int &x, int y) { x = min(x, y); }
  int lowbit(int x) { return x & -x; }
  const int N = 1002000;
  int C[N], n, maxx;
  void Add(int x, int v) {
    while (x <= maxx) {
      smax(C[x], v);
      x += lowbit(x);
    }
  }
  int Query(int x) {
    int ans = 0;
    while (x) {
      smax(ans, C[x]);
      x -= lowbit(x);
    }
    return ans;
  }
  struct Nod {
    int x, y;
  } Node[N];
  bool cmp(Nod a, Nod b) {
    if (a.y != b.y) return a.y < b.y;
    return a.x < b.x;
  }
  signed main() {
    int n = gi();
    for (int i = (int)(1); i <= (int)(n); ++i) {
      Node[i].x = gi();
      Node[i].y = i - Node[i].x;
      smax(maxx, Node[i].x);
    }
    sort(Node + 1, Node + n + 1, cmp);
    int ans = 0;
    for (int i = (int)(1); i <= (int)(n); ++i) {
      if (Node[i].y < 0) continue;
      int g = Query(Node[i].x - 1);
      Add(Node[i].x, g + 1);
      smax(ans, g + 1);
    }
    printf("%d\n", ans);
    return 0;
  }
  