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 N = 200005;
  int t[N * 4];
  struct dt {
    int x, id;
    bool operator<(const dt& a) const {
      return id - x != a.id - a.x ? id - x < a.id - a.x : x < a.x;
    }
  } f[N];
  int n, m;
  inline int ask(int x, int l, int r, int L, int R) {
    if (l == L && r == R) return t[x];
    int mid = (l + r) / 2;
    if (R <= mid) return ask(x * 2, l, mid, L, R);
    if (L > mid) return ask(x * 2 + 1, mid + 1, r, L, R);
    return max(ask(x * 2, l, mid, L, mid),
               ask(x * 2 + 1, mid + 1, r, mid + 1, R));
  }
  inline void modify(int x, int l, int r, int p, int y) {
    t[x] = max(t[x], y);
    if (l == r) return;
    int mid = (l + r) / 2;
    if (p <= mid)
      modify(x * 2, l, mid, p, y);
    else
      modify(x * 2 + 1, mid + 1, r, p, y);
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1, x; i <= n; ++i) {
      scanf("%d", &x);
      if (x <= i) f[++m] = {x, i};
    }
    sort(f + 1, f + 1 + m);
    for (int i = 1; i <= m; ++i) {
      int p = (f[i].x == 1 ? 0 : ask(1, 1, n, 1, f[i].x - 1));
      modify(1, 1, n, f[i].x, p + 1);
    }
    cout << t[1] << endl;
    return 0;
  }
  