http://codeforces.com/contest/1077/problem/E

output

standard output

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:

  • number of problems in each contest is exactly twice as much as in the previous contest (one day ago), the first contest can contain arbitrary number of problems;
  • the total number of problems in all the contests should be maximized.

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.

Input

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.

Output

Print one integer — the maximum number of problems in the set of thematic contests.

Examples
input

Copy
18
2 1 2 10 2 10 10 2 2 1 10 10 10 10 1 1 10 10
output

Copy
14
input

Copy
10
6 6 6 3 6 1000000000 3 3 6 6
output

Copy
9
input

Copy
3
1337 1337 1337
output

Copy
3
Note

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的更多相关文章

  1. Codeforces Round #521 (Div. 3) E. Thematic Contests(思维)

    Codeforces Round #521 (Div. 3)  E. Thematic Contests 题目传送门 题意: 现在有n个题目,每种题目有自己的类型要举办一次考试,考试的原则是每天只有一 ...

  2. Codeforces Round #521 (Div. 3) E. Thematic Contests (离散化,二分)

    题意:有\(n\)个话题,每次都必须选取不同的话题,且话题数必须是上次的两倍,第一次的话题数可以任意,问最多能选取多少话题数. 题解:我们首先用桶来记录不同话题的数量,因为只要求话题的数量,与话题是多 ...

  3. Codeforces Round #521 (Div. 3) D. Cutting Out 【二分+排序】

    任意门:http://codeforces.com/contest/1077/problem/D D. Cutting Out time limit per test 3 seconds memory ...

  4. CodeForces Round #521 (Div.3) D. Cutting Out

    http://codeforces.com/contest/1077/problem/D You are given an array ss consisting of nn integers. Yo ...

  5. Codeforces Round #521 (Div. 3) F1. Pictures with Kittens (easy version)

    F1. Pictures with Kittens (easy version) 题目链接:https://codeforces.com/contest/1077/problem/F1 题意: 给出n ...

  6. CodeForces Round #521 (Div.3) B. Disturbed People

    http://codeforces.com/contest/1077/problem/B There is a house with nn flats situated on the main str ...

  7. CodeForces Round #521 (Div.3) A. Frog Jumping

    http://codeforces.com/contest/1077/problem/A A frog is currently at the point 00 on a coordinate axi ...

  8. Codeforces Round #521 (Div. 3)

    B 题过的有些牵强,浪费了很多时间,这种题一定想好思路和边界条件再打,争取一发过.  D 题最开始读错题,后面最后发现可以重复感觉就没法做了,现在想来,数据量大,但是数据范围小枚举不行,二分还是可以的 ...

  9. Codeforces Round #521 Div. 3 玩耍记

    A:签到. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> ...

随机推荐

  1. MySQL的数据类型(一)

    每一个常量.变量和参数都有数据类型.它用来指定一定的存储格式.约束和有效范围.MySQL提供了多种数据类型.主要有数值型.字符串类型.日期和时间类型.不同的MySQL版本支持的数据类型可能会稍有不同. ...

  2. js函数只触发一次

    如何让js中的函数只被执行一次?我们有时候会有这种需求,即让一个函数只执行一次,第二次调用不会返回任何有价值的值,也不会报错.下面将通过三个小demo展示使用的方法,当做个人笔记. 1.通过闭包来实现 ...

  3. vue.esm.js:578 [Vue warn]: Missing required prop

    问题: 解决: required: true,属性是,这个必须填写

  4. 6 大主流 Web 框架优缺点对比(转)

    英文: Kit Kelly   译文:oschina https://www.oschina.net/translate/web-frameworks-conclusions 是该读些评论和做一些总结 ...

  5. android xml实现animation 4种动画效果

    animation有四种动画类型 分别为alpha(透明的渐变).rotate(旋转).scale(尺寸伸缩).translate(移动),二实现的分发有两种,一种是javaCode,另外一种是XML ...

  6. python 装饰器 (多个装饰器装饰一个函数---装饰器前套一个函数)

    #带参数的装饰器 #500个函数 # import time # FLAGE = False # def timmer_out(flag): # def timmer(func): # def inn ...

  7. C# 设置程序最小化到任务栏右下角,鼠标左键单击还原,右键提示关闭程序

    首先设置程序最小化到任务栏右下角 先给窗口添加一个notifyIcon控件 为notifyIcon控件设置ICO图标(不设置图标将无法在任务栏显示) 给notifyIcon控件添加点击事件 然后是最小 ...

  8. 对bluebird的理解

    前言 Promise:把原来的回调写法分离出来,在异步操作执行完后,用链式调用的方式执行回调函数. 在公众号的开发里面用的const Promise = require('bluebird');con ...

  9. HDU1301 Jungle Roads(Kruskal)

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...

  10. maven中settings文件的配置

    鉴于上一个博客写的maven打包,读者老爷们可能找不到settings文件的配置,这里专门附上settings文件的配置: 有的settings文件只含有一些空标签,需要手动去配置. <?xml ...