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 long long mod = 1e9 + 7;
  const long long maxN = 300005;
  const long long oo = 1e18 + 7;
  long long n, a[maxN];
  long long x, y, z, k;
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    cin >> n;
    vector<pair<long long, long long> > ele;
    for (long long i = 1; i <= n; ++i) {
      cin >> x;
      if (x <= i) {
        ele.push_back(pair<long long, long long>(i - x, x));
      }
    }
    sort(ele.begin(), ele.end());
    vector<long long> lis;
    lis.push_back(-oo);
    long long bonus = 0;
    for (pair<long long, long long> ccc : ele) {
      long long cc = ccc.second;
      if (cc > lis.back())
        lis.push_back(cc);
      else {
        x = lower_bound(lis.begin(), lis.end(), cc) - lis.begin();
        lis[x] = min(lis[x], cc);
      }
    }
    cout << lis.size() - 1 << "\n";
  }
  