Problem:
Score : 600 points

Problem 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.

Constraints

1 \le N \le 3 \times 10^5

1 \le A_i \le N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \ldots A_N

Output

Print N integers.
The t-th (1 \le t \le N) of them should be the answer for the case K=t.

Sample Input 1

3
2 1 2

Sample Output 1

3
1
0

For K = 1, we can do the operation as follows:

Choose the first card to eat.

Choose the second card to eat.

Choose the third card to eat.

For K = 2, we can do the operation as follows:

Choose the first and second cards to eat.

For K = 3, we cannot do the operation at all. Note that we cannot choose the first and third cards at the same time.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

5
2
1
1
1

Sample Input 3

4
1 3 3 3

Sample Output 3

4
1
0
0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
const int maxn = 300000;
int a[maxn], b[maxn], sum[maxn];
bool cmp(const int &A, const int &B)
{
	return A > B;
}
int chkmin(int &A, const int &B)
{
	if (A <= B)
	{
		return 1;
	}
	else
	{
		A = B;
		return 0;
	}
}
int main()
{
	int n;
	scanf("%d", &n);
	for (int k = 1; k <= n; k++)
	{
		scanf("%d", &a[k]);
		b[a[k]]++;
	}
	sort(b + 1, b + n + 1, cmp);
	for (int i = 1; i <= n; i++)
		sum[i] = sum[i - 1] + b[i];
	for (int k = 1; k <= n; k++)
	{
		int ans = n / k;
		int left = 0, right = k - 1;
		while (left <= right)
		{
			int i = (right-left) / 3+left;
			int j = (right-left)*2/3+left;
			if (chkmin(ans, (n - sum[i]) / (k - i)) == 1)
			{
				if (chkmin(ans, (n - sum[j]) / (k - j)) == 1)
				{
					left = i + 1;
				}
				else
				{
					right = j - 1;
				}
			}
			else
			{
				if (chkmin(ans, (n - sum[j]) / (k - j)) == 1)
				{
					left = i + 1;
				}
				else if((n-sum[i])/(k-i) >= (n-sum[j])/(k-j))
				{
					left = i+1;
				}
				else
				{
					right = j-1;
				}
				
			}
			}
			printf("%d\n", ans);
		}
		return 0;
	}