Codeforces Round #384 (Div. 2) 734E Vladik and cards
2 seconds
256 megabytes
standard input
standard output
Vladik was bored on his way home and decided to play the following game. He took n cards and put them in a row in front of himself. Every card has a positive integer number not exceeding 8 written on it. He decided to find the longest subsequence of cards which satisfies the following conditions:
- the number of occurrences of each number from 1 to 8 in the subsequence doesn't differ by more then 1 from the number of occurrences of any other number. Formally, if there are ck cards with number k on them in the subsequence, than for all pairs of integers
the condition |ci - cj| ≤ 1 must hold. - if there is at least one card with number x on it in the subsequence, then all cards with number x in this subsequence must form a continuous segment in it (but not necessarily a continuous segment in the original sequence). For example, the subsequence [1, 1, 2, 2] satisfies this condition while the subsequence [1, 2, 2, 1] doesn't. Note that [1, 1, 2, 2] doesn't satisfy the first condition.
Please help Vladik to find the length of the longest subsequence that satisfies both conditions.
The first line contains single integer n (1 ≤ n ≤ 1000) — the number of cards in Vladik's sequence.
The second line contains the sequence of n positive integers not exceeding 8 — the description of Vladik's sequence.
Print single integer — the length of the longest subsequence of Vladik's sequence that satisfies both conditions.
3
1 1 1
1
8
8 7 6 5 4 3 2 1
8
24
1 8 1 2 8 2 3 8 3 4 8 4 5 8 5 6 8 6 7 8 7 8 8 8
17
In the first sample all the numbers written on the cards are equal, so you can't take more than one card, otherwise you'll violate the first condition.
题意
给定一个序列an,序列中只有1~8的8个整数,让你选出一个子序列,满足下列两个要求
1.不同整数出现的次数相差小于或等于1
2.子序列中整数分布是连续的,即子序列的整数必须是1,1,1....1,2,2,2.....2,2.......连续分布,可以是任意顺序而不要求递增,比如312587644
dp[i][j]记录在j状态下前i个中长度位l个个数 答案就是 ans*l+(8-ans)*(l-1)。
Codeforces Round #384 (Div. 2) 734E Vladik and cards的更多相关文章
- Codeforces Round #384 (Div. 2) E. Vladik and cards 状压dp
E. Vladik and cards 题目链接 http://codeforces.com/contest/743/problem/E 题面 Vladik was bored on his way ...
- Codeforces Round #384 (Div. 2) C. Vladik and fractions 构造题
C. Vladik and fractions 题目链接 http://codeforces.com/contest/743/problem/C 题面 Vladik and Chloe decided ...
- Codeforces Round #384 (Div. 2) A. Vladik and flights 水题
A. Vladik and flights 题目链接 http://codeforces.com/contest/743/problem/A 题面 Vladik is a competitive pr ...
- Codeforces Round #384 (Div. 2) C. Vladik and fractions(构造题)
传送门 Description Vladik and Chloe decided to determine who of them is better at math. Vladik claimed ...
- Codeforces Round #384 (Div. 2) 734E(二分答案+状态压缩DP)
题目大意 给定一个序列an,序列中只有1~8的8个整数,让你选出一个子序列,满足下列两个要求 1.不同整数出现的次数相差小于等于1 2.子序列中整数分布是连续的,即子序列的整数必须是1,1,1.... ...
- queue+模拟 Codeforces Round #304 (Div. 2) C. Soldier and Cards
题目传送门 /* 题意:两堆牌,每次拿出上面的牌做比较,大的一方收走两张牌,直到一方没有牌 queue容器:模拟上述过程,当次数达到最大值时判断为-1 */ #include <cstdio&g ...
- Codeforces Round #384 (Div. 2) //复习状压... 罚时爆炸 BOOM _DONE
不想欠题了..... 多打打CF才知道自己智商不足啊... A. Vladik and flights 给你一个01串 相同之间随便飞 没有费用 不同的飞需要费用为 abs i-j 真是题意杀啊, ...
- Codeforces Round #384 (Div. 2)D - Chloe and pleasant prizes 树形dp
D - Chloe and pleasant prizes 链接 http://codeforces.com/contest/743/problem/D 题面 Generous sponsors of ...
- Codeforces Round #384 (Div. 2)B. Chloe and the sequence 数学
B. Chloe and the sequence 题目链接 http://codeforces.com/contest/743/problem/B 题面 Chloe, the same as Vla ...
随机推荐
- 解决mysql Table ‘xxx’ is marked as crashed and should be repaired的问题。
解决mysql Table 'xxx' is marked as crashed and should be repaired的问题. 某个表在进行数据插入和更新时突然出现Table 'xxx' is ...
- mysql nonInstall 版本的安装与配置
最近用到mysql,发现如果想使用最新版本64 bit mysql 需要独特的配置和使用方式 结合最近的研究总结一下安装过程. 首先下载:http://dev.mysql.com/downloads/ ...
- ztree-demo 2
<!DOCTYPE html> <HTML> <HEAD> <TITLE> ZTREE DEMO - Async</TITLE> <m ...
- zookeeper3.3.6 伪分布式安装
下载地址(http://zookeeper.apache.org/releases.html#download) 一:下载zookeeper的安装包,解压,进入到zk的目录文件,进入conf目录 ...
- java十进制转十六进制
package com.ds.detect; import java.util.Scanner; public class ToHEX{ public static void main(String[ ...
- python之路:Day02 --- Python基础2
本节内容 1.列表操作 2.元组操作 3.字符串操作 4.字典操作 5.集合操作 6.文件操作 7.字符编码与转换 一.列表操作 定义列表 names = ['Ming',"Hua" ...
- Redsi和Memcached区别总结
首先谈谈Redis和Memcached它们都是缓存在内存中的,唯一的区别就是Redis它本身会周期性的把 更新的一些数据写入到磁盘或者修改操作写入追加的记录文件中,并且在此基础上实现master-sl ...
- 《C#本质论》读书笔记(18)多线程处理
.NET Framework 4.0 看(本质论第3版) .NET Framework 4.5 看(本质论第4版) .NET 4.0为多线程引入了两组新API:TPL(Task Parallel Li ...
- centos 6.6 系统中配置sendmail和dovecot
网上介绍sendmail的文章千百种,很少有跟着做下来一次成功的.多少都有些说的不准确的地方. 我给大家共享一下我经过实验环境测试,完全可行的方法. 1.软件准备 操作系统:centos6.6我选择c ...
- “Operation is not valid due to the current state of the object.”
将Repeater单页显示的2000条数据一次性提交的时候出现这个错误: Operation is not valid due to the current state of the object. ...