标签：lower ble thml seve wan for host tle data-
http://codeforces.com/contest/1077/problem/E
Polycarp has prepared nn competitive programming problems. The topic of the ii-th problem is aiai, and some problems‘ topics may coincide.
Polycarp has to host several thematic contests. All problems in each contest should have the same topic, and all contests should have pairwise distinct topics. He may not use all the problems. It is possible that there are no contests for some topics.
Polycarp wants to host competitions on consecutive days, one contest per day. Polycarp wants to host a set of contests in such a way that:
Your task is to calculate the maximum number of problems in the set of thematic contests. Note, that you should not maximize the number of contests.
The first line of the input contains one integer nn (1≤n≤2?1051≤n≤2?105) — the number of problems Polycarp has prepared.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) where aiai is the topic of the ii-th problem.
Print one integer — the maximum number of problems in the set of thematic contests.
18 2 1 2 10 2 10 10 2 2 1 10 10 10 10 1 1 10 10
14
10 6 6 6 3 6 1000000000 3 3 6 6
9
3 1337 1337 1337
3
In the first example the optimal sequence of contests is: 22 problems of the topic 11, 44 problems of the topic 22, 88 problems of the topic 1010.
In the second example the optimal sequence of contests is: 33 problems of the topic 33, 66 problems of the topic 66.
In the third example you can take all the problems with the topic 13371337 (the number of such problems is 33 so the answer is 33) and host a single contest.
代码：
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const int N = 1e6 + 5;
int n;
int a[N];
int main() {
scanf("%d", &n);
int cnt = 0;
map<int, int> mp;
for(int i = 1; i <= n; i ++) {
int x;
scanf("%d", &x);
if(!mp[x]) {
cnt ++;
mp[x] = cnt;
}
a[mp[x]] ++;
}
sort(a + 1, a + 1 + cnt);
int ans = 0;
for(int i = 1; i <= n; i ++) {
int st = 1;
int m = 0;
for(int k = i; k <= n; k *= 2) {
int pos = lower_bound(a + st, a + 1 + cnt, k) - a;
if(pos == cnt + 1) break;
m += k;
st = pos + 1;
}
ans = max(ans, m);
}
printf("%d\n", ans);
return 0;
}
CodeForces Round #521 (Div.3) E. Thematic Contests
标签：lower ble thml seve wan for host tle data-
原文地址：https://www.cnblogs.com/zlrrrr/p/10085595.html
