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;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;
inline ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
inline ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }
template<class S,class T> ostream& operator << (ostream& out,const pair<S,T>& o){ out << "(" << o.first << "," << o.second << ")"; return out; }
template<class T> ostream& operator << (ostream& out,const vector<T>& V){ for(int i = 0; i < V.size(); i++){ out << V[i]; if(i!=V.size()-1) out << " ";} return out; }
template<class T> ostream& operator << (ostream& out,const vector<vector<T> >& Mat){ for(int i = 0; i < Mat.size(); i++) { if(i != 0) out << endl; out << Mat[i];} return out; }
template<class S,class T> ostream& operator << (ostream& out,const map<S,T>& mp){ out << "{ "; for(auto it = mp.begin(); it != mp.end(); it++){ out << it->first << ":" << it->second; if(mp.size()-1 != distance(mp.begin(),it)) out << ", "; } out << " }"; return out; }
template<typename T>vector<T> make_v(size_t a){return vector<T>(a);}
template<typename T,typename... Ts>auto make_v(size_t a,Ts... ts){return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));}
template<typename T,typename V> typename enable_if<is_class<T>::value==0>::type fill_v(T &t,const V &v){t=v;}
template<typename T,typename V> typename enable_if<is_class<T>::value!=0>::type fill_v(T &t,const V &v){for(auto &e:t) fill_v(e,v);}
/*
 <url:https://atcoder.jp/contests/abc143/tasks/abc143_f>
 問題文============================================================
 =================================================================
 解説=============================================================
 ================================================================
 */

template<class Type>
Type solve(Type res = Type()){
    map<int,int> AtoCnt;
    int N; cin >> N;
    for(int i = 0; i < N;i++){
        int A; cin >> A;
        AtoCnt[A]++;
    }
    map<int,int> CntNum;
    for(auto p:AtoCnt){
        CntNum[p.second]++;
    }
    vector<pii> C;
    for(auto p:CntNum) C.push_back(p);
    sort(C.rbegin(),C.rend()); C.push_back(pii(0,0));
    int n = (int)C.size();
    
    // cout << C << endl;
    for(int K = 1; K <= N; K++){
        int ans = 0;
        priority_queue<pii> pq;
        int y = 0;
        for(int i = 1; i < n;i++){
            int dx = C[i-1].first - C[i].first;
            y += C[i-1].second;
            
            if(y < K){
                pq.push(pii(y,dx));
            }else{
                ans += y/K * dx;
                
                int L = (y*dx) - (y/K*K * dx);
                // cout << y << " " << dx << " " << K << endl;
                while(L && pq.size()){
                    pii p = pq.top(); pq.pop();
                    
                    int ness = K - p.first;
                    int get = min(L/ness,p.second);
                    // cout << p << endl;
                    // cout << L << " " << ness << " " << get << endl;
                    p.second -= get;
                    ans += get;
                    L = L - ness*get;
                    if(p.second == 0) continue;
                    
                    pq.push(pii(p.first+L,1));
                    if(p.second-1) pq.push(pii(p.first,p.second-1));
                }
                
                if(L){
                    pq.push(pii(L,1));
                }
            }
        }
        cout << ans << endl;
    }
    return res;
}
int main(void) {
    cin.tie(0); ios_base::sync_with_stdio(false);
    solve(0);
    // cout << fixed << setprecision(15) << solve<ll>() << endl;
    return 0;
}