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;
  struct node {
    int a, c;
  } a[200005];
  int n, dp[200005], maxlen;
  bool cmp(node a, node b) { return a.a < b.a || a.a == b.a && a.c > b.c; }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i].a);
    for (int i = 1; i <= n; i++) a[i].c = i - a[i].a;
    sort(a + 1, a + 1 + n, cmp);
    dp[0] = -1e9;
    for (int i = 1; i <= n; i++)
      if (a[i].c >= 0) {
        if (dp[maxlen] <= a[i].c) {
          dp[++maxlen] = a[i].c;
        } else {
          int k = upper_bound(dp, dp + 1 + maxlen, a[i].c) - dp;
          dp[k] = a[i].c;
        }
      }
    printf("%d\n", maxlen);
    return 0;
  }
  