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 maxn = 5e5 + 7;
  int a[maxn], b[maxn], c[maxn];
  struct segtree {
    int l, r, mid;
    long long val = 0;
    segtree *lc, *rc;
    segtree(int L, int R) {
      l = L;
      r = R;
      mid = (l + r) / 2;
      if (l != r) {
        lc = new segtree(l, mid);
        rc = new segtree(mid + 1, r);
      }
    }
    void apply(long long X) { val = max(val, X); }
    void update(int L, int R, long long X) {
      if (l == L && r == R) {
        apply(X);
        return;
      } else if (R <= mid)
        lc->update(L, R, X);
      else if (L >= mid + 1)
        rc->update(L, R, X);
      else {
        lc->update(L, mid, X);
        rc->update(mid + 1, R, X);
      }
      val = max(lc->val, rc->val);
    }
    long long query(int L, int R) {
      if (l == L && r == R)
        return val;
      else if (R <= mid)
        return lc->query(L, R);
      else if (L >= mid + 1)
        return rc->query(L, R);
      else {
        return max(lc->query(L, mid), rc->query(mid + 1, R));
      }
    }
  } * segroot;
  vector<pair<int, int> > pos[maxn];
  long long dp[maxn];
  int main() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) {
      b[i] = i - a[i];
    }
    for (int i = 1; i <= n; ++i) {
      c[i] = i - b[i];
      if (b[i] >= 0) pos[b[i]].push_back({c[i], i});
    }
    for (int i = 0; i <= 200000; ++i) {
      if (pos[i].size()) sort(pos[i].begin(), pos[i].end());
    }
    segroot = new segtree(0, 200001);
    for (int i = 0; i <= 200000; ++i) {
      if (pos[i].size() == 0) continue;
      for (auto p : pos[i]) {
        int v = p.second;
        int now = c[v];
        dp[v] = segroot->query(0, now - 1) + 1;
        segroot->update(now, now, dp[v]);
      }
    }
    long long ans = 0;
    for (int i = 1; i <= n; ++i) ans = max(ans, dp[i]);
    cout << ans << endl;
  }
  