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 LIS(vector<long long>& v) {
    long long n = v.size();
    if (!n) return 0;
    vector<long long> dp;
    dp = {v[0]};
    for (long long i = 1; i < n; ++i) {
      auto it = lower_bound(dp.begin(), dp.end(), v[i]);
      if (it == dp.end())
        dp.push_back(v[i]);
      else
        *it = v[i];
    }
    return dp.size();
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    long long n;
    cin >> n;
    vector<pair<long long, long long>> a;
    for (long long i = 1; i <= n; ++i) {
      long long x;
      cin >> x;
      if (i >= x) {
        a.push_back({i - x, x});
      }
    }
    sort(a.begin(), a.end());
    vector<long long> b;
    for (auto& [i, j] : a) {
      b.push_back(j);
    }
    cout << LIS(b);
    return 0;
  }
  