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;
  long long read() {
    long long x = 0, f = 1;
    char ch = ' ';
    while (!isdigit(ch)) {
      ch = getchar();
      if (ch == '-') f = -1;
    }
    while (isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
  }
  const int N = 2e5;
  int n, f[200005], c[200005], ans;
  void jia(int i, int x) {
    while (i <= 200001) {
      c[i] = max(c[i], x);
      i += i & -i;
    }
  }
  int he(int x) {
    int s = 0;
    while (x) {
      s = max(s, c[x]);
      x -= x & -x;
    }
    return s;
  }
  pair<int, int> a[200005];
  int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      int x = read();
      a[i] = {i - x, x};
    }
    sort(a + 1, a + n + 1);
    jia(1, 1);
    for (int i = 1; i <= n; i++) {
      if (a[i].first < 0) continue;
      f[i] = he(a[i].second);
      jia(a[i].second + 1, f[i] + 1);
      ans = max(ans, f[i]);
    }
    cout << ans;
  }
  