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;
  inline int read() {
    int k = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      k = k * 10 + ch - '0';
      ch = getchar();
    }
    return k * f;
  }
  inline void write(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
  }
  inline void writeln(int x) {
    write(x);
    puts("");
  }
  const int N = 4e5 + 10, M = 2e5;
  int n;
  int a[N], cntt = 0;
  int f[N], maxx = 0;
  struct ppap {
    int x, v;
  } b[N];
  inline bool cmp(ppap a, ppap b) { return a.v == b.v ? a.x > b.x : a.v < b.v; }
  inline void update(int x, int v) {
    for (; x <= M + n; x += x & -x) f[x] = max(f[x], v);
  }
  inline int smax(int x) {
    int ans = -1;
    for (; x; x -= x & -x) {
      ans = max(ans, f[x]);
    }
    return ans;
  }
  int main() {
    n = read();
    memset(f, -1, sizeof f);
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      b[i] = (ppap){i, a[i]};
      maxx = max(maxx, a[i]);
    }
    sort(b + 1, b + n + 1, cmp);
    update(0 + M, 0);
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      int x = b[i].x, y = b[i].v;
      int p = x - y;
      int v = smax(p + M) + 1;
      if (v == 0 && p < 0) v = -1;
      ans = max(ans, v);
      update(p + M, v);
    }
    writeln(ans);
    return 0;
  }
  