#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (b); i >= (a); i--)
#define SZ(x) ((int)x.size())
#define all(x) x.begin(), x.end()
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define st first
#define nd second
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
auto &operator<<(auto &o, pair<auto, auto> p) {
return o << "(" << p.st << ", " << p.nd << ")";
}
auto operator<<(auto &o, auto x)->decltype(end(x), o) {
o << "{"; int i=0; for(auto e: x) o << ", " + 2*!i++ << e;
return o << "}";
}
#define deb(x...) cerr << "[" #x "]: ", [](auto...$) { ((cerr<<$<<"; "),...) << endl; }(x)
const int nax = 3e5 + 5;
int n;
map<int, int> cnt;
int ans[nax];
void solve(){
cin >> n;
rep(i, 1, n){
int x; cin >> x;
cnt[x] += 1;
}
for(auto [_, c] : cnt){
rep(i, 1, c){
ans[i] += c / i * i;
}
}
rep(i, 1, n) cout << ans[i] << " ";
cout << "\n";
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int tt = 1;
// cin >> tt;
rep(te, 1, tt) solve();
return 0;
}
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 | #pragma GCC optimize ("O3") #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i = (a); i <= (b); i++) #define per(i, a, b) for (int i = (b); i >= (a); i--) #define SZ(x) ((int)x.size()) #define all(x) x.begin(), x.end() #define pb push_back #define mp make_pair #define mt make_tuple #define st first #define nd second using ll = long long; using vi = vector<int>; using pii = pair<int, int>; using pll = pair<ll, ll>; auto &operator<<(auto &o, pair<auto, auto> p) { return o << "(" << p.st << ", " << p.nd << ")"; } auto operator<<(auto &o, auto x)->decltype(end(x), o) { o << "{"; int i=0; for(auto e: x) o << ", " + 2*!i++ << e; return o << "}"; } #define deb(x...) cerr << "[" #x "]: ", [](auto...$) { ((cerr<<$<<"; "),...) << endl; }(x) const int nax = 3e5 + 5; int n; map<int, int> cnt; int ans[nax]; void solve(){ cin >> n; rep(i, 1, n){ int x; cin >> x; cnt[x] += 1; } for(auto [_, c] : cnt){ rep(i, 1, c){ ans[i] += c / i * i; } } rep(i, 1, n) cout << ans[i] << " "; cout << "\n"; } int main() { ios::sync_with_stdio(0); cin.tie(0); int tt = 1; // cin >> tt; rep(te, 1, tt) solve(); return 0; } |
English