output_description: Print N integers. The t-th (1 \le t \le N) of them should be the answer for
  the case K=t.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      A_1 A_2 \ldots A_N
attempt: import collections
  import numpy as np
  
  # リストarrからx個の同じ長さの列を取り出す時の最大の長さ
  # 発想: x個取り出す時、ある文字をx個より多く取り出して列に含むことはない
  #       一方ある文字をその文字がある数よりも取り出して使うこともできない
  #       つまり、min(arr.count(a), x)で、x個の同じ長さの列に使えるaの数がわかる
  #       さらに、sum(min(arr.count(a),x))で、x個の同じ長さの列に使える文字の数がわかる
  #       よって、sum(min(arr.count(a),x))//xで、x個の同じ長さの列の長さがわかる
  #       発展的に、longest(arr,x)までの計算状況からlongest(arr,x+1)が計算できる
  # def longest(arr, x):    # legacy
  #     return sum(min(arr.count(a), x) for a in arr) // x
  
  
  n = int(input())
  a = list(map(int, input().split()))
  
  counter = collections.Counter(a)  # O(n)
  nums = list(counter.values())  # O(n)
  num_counter = collections.Counter(nums)  # O(n)
  lastest_add = np.count_nonzero(nums)  # O(n)
  building, longests = [0], [0]
  for i in range(1, n + 1):  # O(n)
      appended = building[-1] + lastest_add  # O(1)
      building.append(appended)  # O(1)
      longests.append(appended // i)  # O(1)
      lastest_add -= num_counter[i]  # O(1)
  
  
  lastest_longest = 0
  k_most = []
  for i, x in enumerate(reversed(longests)):
      while x > lastest_longest:
          lastest_longest += 1
          k_most.append(n - i)
  
  k_most.extend([0] * (n - len(k_most)))  # zero padding
  
  for longests_num in k_most:
      print(longests_num)
  
  # print('----------------')
  # for k in range(1, n+1):
  #     maxindexes = list(i for i, x in enumerate(longests) if x >= k)
  #     if maxindexes:
  #         print(maxindexes[-1])
  #     else:
  #         print(0)
  
problem_description: Statement
  
  Takahashi has N cards. The i-th of these cards has an integer A_i written on
  it.
  
  Takahashi will choose an integer K, and then repeat the following operation
  some number of times:
  
    * Choose exactly K cards such that the integers written on them are all different, and eat those cards. (The eaten cards disappear.)
  
  For each K = 1,2, \ldots, N, find the maximum number of times Takahashi can do
  the operation.
samples: [{"input": "3\n    2 1 2", "output": "3\n    1\n    0\n    \n\nFor K = 1, we can do the operation as follows:\n\n  * Choose the first card to eat.\n  * Choose the second card to eat.\n  * Choose the third card to eat.\n\nFor K = 2, we can do the operation as follows:\n\n  * Choose the first and second cards to eat.\n\nFor K = 3, we cannot do the operation at all. Note that we cannot choose the\nfirst and third cards at the same time.\n\n* * *"}, {"input": "5\n    1 2 3 4 5", "output": "5\n    2\n    1\n    1\n    1\n    \n\n* * *"}, {"input": "4\n    1 3 3 3", "output": "4\n    1\n    0\n    0"}]