Codeforces Round #544 (Div. 3) Editorial C. Balanced Team
http://codeforces.com/contest/1133/problem/C
time limit per test 2 seconds
memory limit per test 256 megabytes
inputstandard input
outputstandard output
You are a coach at your local university. There are $n$ students under your supervision, the programming skill of the i-th student is $a_i$.
You have to create a team for a new programming competition. As you know, the more students some team has the more probable its victory is! So you have to create a team with the maximum number of students. But you also know that a team should be balanced. It means that the programming skill of each pair of students in a created team should differ by no more than 5.
Your task is to report the maximum possible number of students in a balanced team.
Input
The first line of the input contains one integer $n$($1≤n≤2⋅10^5$) — the number of students.
The second line of the input contains $n$ integers $a_1,a_2,…,a_n$($1≤a_i≤10^9$), where $a_i$ is a programming skill of the i-th student.
Output
Print one integer — the maximum possible number of students in a balanced team.
Examples
input
6
1 10 17 12 15 2
output
3
input
10
1337 1337 1337 1337 1337 1337 1337 1337 1337 1337
output
10
input
6
1 1000 10000 10 100 1000000000
output
1
Note
In the first example you can create a team with skills [12,17,15].
In the second example you can take all students in a team because their programming skills are equal.
In the third example you can create a team consisting of a single student (and you cannot create a team consisting of at least two students).
解题思路
官方:Let's sort all values in non-decreasing order. Then we can use two pointers to calculate for each student
Codeforces Round #544 (Div. 3) Editorial C. Balanced Team的更多相关文章
- Codeforces Round #544 (Div. 3) E. K Balanced Teams (DP)
题意:有\(n\)个人,每个人的能力值是\(a_i\),现在你想将这些人分成\(k\)组(没必要全选),但是每组中最高水平和最低水平的人的能力差值必须\(\le 5\),问最多能选多少人. 题解:想了 ...
- Codeforces Round #544 (Div. 3) 题解
Codeforces Round #544 (Div. 3) D. Zero Quantity Maximization 题目链接:https://codeforces.com/contest/113 ...
- Codeforces Round #590 (Div. 3) Editorial
Codeforces Round #590 (Div. 3) Editorial 题目链接 官方题解 不要因为走得太远,就忘记为什么出发! Problem A 题目大意:商店有n件商品,每件商品有不同 ...
- Codeforces Round #747 (Div. 2) Editorial
Codeforces Round #747 (Div. 2) A. Consecutive Sum Riddle 思路分析: 一开始想起了那个公式\(l + (l + 1) + - + (r − 1) ...
- CodeForces Round #544 Div.3
A. Middle of the Contest 代码: #include <bits/stdc++.h> using namespace std; int h1, m1, h2, m2; ...
- Codeforces Round #544 (Div. 3)解题报告
A.Middle of the Contest 考虑把输入的时间单位化成分钟,相加除以2就好了 #include<bits/stdc++.h> using namespace std; # ...
- Codeforces Round #544 (Div. 3) C. Balanced Team
链接:https://codeforces.com/contest/1133/problem/C 题意: 给n个数, 在这n个数中选最多n个数,来组成一个队伍. 保证这n个数的最大最小差值不大于5. ...
- Codeforces Round #544 (Div. 3) dp + 双指针
https://codeforces.com/contest/1133/problem/E 题意 给你n个数(n<=5000),你需要对其挑选并进行分组,总组数不能超过k(k<=5000) ...
- Codeforces Round #544 (Div. 3) D. Zero Quantity Maximization
链接:https://codeforces.com/contest/1133/problem/D 题意: 给两个数组a,b. 同时ci = ai * d + bi. 找到一个d使c数组中的0最多. 求 ...
随机推荐
- bzoj 1633: [Usaco2007 Feb]The Cow Lexicon 牛的词典【dp】
预处理出g[i][j]表示原串第i个匹配第j个单词需要去掉几个字母(匹配不上为-1) 设f[i]为i及之后满足条件要去掉的最少字母 倒着dp! f[i]初始为f[i+1]+1,转移方程为f[i]=mi ...
- $CF1153A\ Serval\ and\ Bus$
看大佬的代码都好复杂(不愧是大佬\(orz\) 蒟蒻提供一种思路 因为求的是最近的车对吧\(qwq\) 所以我们可以用一个\(while\)循环所以没必要去用什么 \(for...\) 至于这是\(d ...
- [转]C语言文件操作函数大全(超详细)
fopen(打开文件)相关函数 open,fclose表头文件 #include<stdio.h>定义函数 FILE * fopen(const char * path,const cha ...
- 转 DOS(CMD)下批处理换行问题/命令行参数换行 arg ms-dos
DOS(CMD)下批处理换行问题本人经常写一些DOS批处理文件,由于批处理中命令的参考较多且长,写在一行太不容易分辨,所以总想找个办法把一条命令分行来写,今天终于试成功两种方法.一.在CMD下,可以用 ...
- 297 Serialize and Deserialize Binary Tree 二叉树的序列化与反序列化
序列化是将一个数据结构或者对象转换为连续的比特位的操作,进而可以将转换后的数据存储在一个文件或者内存中,同时也可以通过网络传输到另一个计算机环境,采取相反方式重构得到原数据.请设计一个算法来实现二叉树 ...
- Android PopupWindow使用时注意
项目中使用PopupWindown出现的坑 1.部分设备,PopWindow在Android4.0后版本,出现NullPointerException调用以下方法可解决, fixPopupWindow ...
- 联想 S5 Pro GT(L78091)免解锁BL 免rec 保数据 ROOT Magisk Xposed 救砖 ZUI5.0.047
>>>重点介绍<<< 第一:本刷机包可卡刷可线刷,刷机包比较大的原因是采用同时兼容卡刷和线刷的格式,所以比较大第二:[卡刷方法]卡刷不要解压刷机包,直接传入手机后用 ...
- 关于c++11中static类对象构造函数线程安全的验证
在c++11中,static静态类对象在执行构造函数进行初始化的过程是线程安全的,有了这个特征,我们可以自己动手轻松的实现单例类,关于如何实现线程安全的单例类,请查看c++:自己动手实现线程安全的c+ ...
- react Native环境 搭建
react Native的优点:跨平台 低投入高回报 性能高 支持动态更新.一才两用(ios和Android) 开发成本第 代码复用率高.windows环境搭建react Native开发环境1.安装 ...
- 3星|《OKR:源于英特热和谷歌的目标管理利器》:OKR原理、实施手册、实施过的公司的访谈
OKR原理与实施手册,另外附了几家实施过OKR的公司的访谈. 书中表述的OKR思想,我认为是这两点: 1:始终聚焦在最重要的目标上: 2:不以OKR为考核员工的指标: Kindle电子版排版有小缺陷, ...