CodeForces Round #521 (Div.3) E. Thematic Contests
http://codeforces.com/contest/1077/problem/E
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.
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的更多相关文章
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- Codeforces Round #521 Div. 3 玩耍记
A:签到. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> ...
随机推荐
- php第一节(入门语法、数据类型)
<?php /** * 变量命名用 $ 符 * 变量名称的命名规范 * 1.变量名称以$标示 * 2.变量名称只能以字母和下划线开头 * 3.变量的名称只能包含字母.下划线.数字 * 4.变量名 ...
- BZOJ1569: [JSOI2008]Blue Mary的职员分配(dp 暴力)
Time Limit: 15 Sec Memory Limit: 162 MBSubmit: 483 Solved: 189[Submit][Status][Discuss] Descriptio ...
- ABAP术语-Lock Mode
Lock Mode 原文:http://www.cnblogs.com/qiangsheng/archive/2008/02/29/1085732.html Status that determine ...
- go 下面定义嵌套结构
package main import ( "fmt" ) const ( URL = "http://www.163.com" UID = "adm ...
- PyCharm入门第一步-——创建并运行第一个Python项目
创建项目 点击Create New Project 创建项目 输入自己的项目名,点击Create创建 创建文件 右键项目名创建python文件 创建一个HelloPython文件 输入print(&q ...
- Mysql 5.7 开启远程连接
1 在控制台执行 mysql -uroot -p 系统提示输入数据库root用户的密码,输入完成后即进入mysql控制台 2 选择数据库 mysql -uroot -p use mysql; 开启远程 ...
- JAVA 泛型方法<T>
public static void main(String[] args) throws Exception { String[] arr = new String[]{"1", ...
- Python3 使用基本循环实现多级目录(思路)
一.多级目录设计: 1. 通过循环的方式显示菜单和进入菜单 2. 设置标志位以提供回退上一层菜单 2. 设置标志位以提供退出程序 二.注意要点: 1. 菜单样式,层次关系不要弄混乱 2. 当输入错误时 ...
- 查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理(001),马克思(002),OO&UML (003),数据库(004)
SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企 ...
- JAVA大作业汇总2
JAVA大作业2 代码 package thegreatwork; //Enum一般用来表示一组相同类型的常量,这里用于表示运动方向的枚举型常量,每个方向对象包括方向向量. public enum D ...