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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)

#![allow(unused_imports)]
#![allow(dead_code)]
#![allow(non_camel_case_types)]
#![allow(non_snake_case)]

use std::cmp::*;
use std::collections::*;
use std::ops::*;
use std::io::{Write, BufWriter};

// static MOD: usize = 998244353;
static MOD: usize = 1000000007;

// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, [ $t:tt ]) => {{
        let len = read_value!($next, usize);
        (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    }};

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

macro_rules! println_stderr(
    ($($arg:tt)*) => { {
        let r = writeln!(&mut ::std::io::stderr(), $($arg)*);
        r.expect("failed printing to stderr");
    } }
);

fn solve() {
    input! {
        n: usize,
        a: [i32; n],
    }  
    let mut b = a.clone();
    b.sort();
    let mut sum = 1;
    let mut v = vec![];
    for i in 1..n {
        if b[i]==b[i-1] {
            sum += 1;
        }
        else{
            v.push(sum);
            sum = 1;
        }
    }
    if sum!=0 {v.push(sum);}
    let ma = v.iter().max().unwrap();
    for k in 1..n+1 {
        if *ma <= n/k {
            println!("{}", n/k);
            continue;
        }
        if v.len() < k {
            println!("{}", 0);
            continue;
        }
        let mut heap = BinaryHeap::new();
        for &i in &v {
            heap.push(i);
        }
        let mut ans = 0;
        while heap.len()>=k {
            ans += 1;
            let mut tmp = vec![];
            for j in 0..k {
                let val = heap.pop().unwrap()-1;
                if val > 0 {tmp.push(val);}
            }
            for &j in &tmp {
                heap.push(j);
            }
        }
        println!("{}", ans);
    }
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}