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;
  int n;
  int a[200007];
  int c[1000007];
  vector<pair<int, int> > p;
  void add(int v, int val) {
    v += 200000;
    for (int i = v; i <= 400007; i += ((i) & (-i))) {
      c[i] = max(c[i], val);
    }
    return;
  }
  int ask(int v) {
    v += 200000;
    int res = 0;
    for (int i = v; i; i -= ((i) & (-i))) {
      res = max(res, c[i]);
    }
    return res;
  }
  vector<int> te[200007];
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (a[i] > n || a[i] > i) continue;
      p.emplace_back(a[i], i - a[i]);
    }
    int ans = 0;
    sort(p.begin(), p.end());
    for (auto [x, y] : p) {
      te[x].push_back(y);
    }
    for (int i = 1; i <= n; i++) {
      vector<pair<int, int> > q;
      for (auto y : te[i]) {
        int tp = ask(y) + 1;
        ans = max(ans, tp);
        q.emplace_back(y, tp);
      }
      for (auto [y, tp] : q) {
        add(y, tp);
      }
    }
    printf("%d\n", ans);
    return 0;
  }
  