C. Brutality
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are playing a new famous fighting game: Kortal Mombat XII. You have to perform a brutality on your opponent's character.

You are playing the game on the new generation console so your gamepad have 2626 buttons. Each button has a single lowercase Latin letter from 'a' to 'z' written on it. All the letters on buttons are pairwise distinct.

You are given a sequence of hits, the ii-th hit deals aiai units of damage to the opponent's character. To perform the ii-th hit you have to press the button sisi on your gamepad. Hits are numbered from 11 to nn.

You know that if you press some button more than kk times in a row then it'll break. You cherish your gamepad and don't want to break any of its buttons.

To perform a brutality you have to land some of the hits of the given sequence. You are allowed to skip any of them, however changing the initial order of the sequence is prohibited. The total damage dealt is the sum of aiai over all ii for the hits which weren't skipped.

Note that if you skip the hit then the counter of consecutive presses the button won't reset.

Your task is to skip some hits to deal the maximum possible total damage to the opponent's character and not break your gamepad buttons.

Input

The first line of the input contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of hits and the maximum number of times you can push the same button in a row.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109), where aiai is the damage of the ii-th hit.

The third line of the input contains the string ss consisting of exactly nn lowercase Latin letters — the sequence of hits (each character is the letter on the button you need to press to perform the corresponding hit).

Output

Print one integer dmgdmg — the maximum possible damage to the opponent's character you can deal without breaking your gamepad buttons.

Examples
input

Copy
7 3
1 5 16 18 7 2 10
baaaaca
output

Copy
54
input

Copy
5 5
2 4 1 3 1000
aaaaa
output

Copy
1010
input

Copy
5 4
2 4 1 3 1000
aaaaa
output

Copy
1009
input

Copy
8 1
10 15 2 1 4 8 15 16
qqwweerr
output

Copy
41
input

Copy
6 3
14 18 9 19 2 15
cccccc
output

Copy
52
input

Copy
2 1
10 10
qq
output

Copy
10
Note

In the first example you can choose hits with numbers [1,3,4,5,6,7][1,3,4,5,6,7] with the total damage 1+16+18+7+2+10=541+16+18+7+2+10=54.

In the second example you can choose all hits so the total damage is 2+4+1+3+1000=10102+4+1+3+1000=1010.

In the third example you can choose all hits expect the third one so the total damage is 2+4+3+1000=10092+4+3+1000=1009.

In the fourth example you can choose hits with numbers [2,3,6,8][2,3,6,8]. Only this way you can reach the maximum total damage 15+2+8+16=4115+2+8+16=41.

In the fifth example you can choose only hits with numbers [2,4,6][2,4,6] with the total damage 18+19+15=5218+19+15=52.

In the sixth example you can change either first hit or the second hit (it does not matter) with the total damage 1010.

题目大意:

给你一个数字n和k,n代表有n次操作,k代表一个按钮最大连续按的次数。

接下来就是一行数字,代表这一次的伤害值,

接下来又是一串字母,这个代表第i次操作要按的按钮,这个与上面的相对应,让你求最大的伤害值。

这个就是一个贪心的题目,需要你仔细一点,这个题目读懂就好写了,这个就是如果要连续按下同一个按钮就超过k次这个按钮就错了,

但是呢,如果中途有别的按钮按了,这个按钮之前的次数就会消除。

C. Brutality Educational Codeforces Round 59 (Rated for Div. 2) 贪心+思维的更多相关文章

  1. Educational Codeforces Round 59 (Rated for Div. 2) DE题解

    Educational Codeforces Round 59 (Rated for Div. 2) D. Compression 题目链接:https://codeforces.com/contes ...

  2. Educational Codeforces Round 59 (Rated for Div. 2) (前四题)

    A. Digits Sequence Dividing(英文速读) 练习英语速读的题,我还上来昏迷一次....只要长度大于2那么一定可以等于2那么前面大于后面就行其他no 大于2的时候分成前面1个剩下 ...

  3. Educational Codeforces Round 59 (Rated for Div. 2) E 区间dp + 状态定义 + dp预处理(分步dp)

    https://codeforces.com/contest/1107/problem/E 题意 给出01字符串s(n<=100),相邻且相同的字符可以同时消去,一次性消去i个字符的分数是\(a ...

  4. Educational Codeforces Round 59 (Rated for Div. 2)

    熬夜爆肝,智商急剧下降 坐标UTC+8晚上23:35开始 晚上脑袋转的慢,非常慢 T1上来先做还花了好几分钟 T2本来是有式子的我TM写数位DP写炸了然后才发现是有公式 T3英语不好,一开始题意没读懂 ...

  5. 【考试记录】Educational Codeforces Round 59 (Rated for Div. 2)

    本来准备划水,结果被垃圾题艹翻了…… T2题意: 定义一个数$x$的数字根$S(x)$为:将其各位数字相加得到一个新数,再将新数的数字和相加直到得到一个个位数,就是该数的数字根. 例如:$S(38)= ...

  6. C. Playlist Educational Codeforces Round 62 (Rated for Div. 2) 贪心+优先队列

    C. Playlist time limit per test 2 seconds memory limit per test 256 megabytes input standard input o ...

  7. Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship

    Problem   Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec P ...

  8. Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems(动态规划+矩阵快速幂)

    Problem   Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec P ...

  9. Educational Codeforces Round 43 (Rated for Div. 2)

    Educational Codeforces Round 43 (Rated for Div. 2) https://codeforces.com/contest/976 A #include< ...

随机推荐

  1. JPA、Hibernate、Spring data jpa之间的关系,终于明白了

    什么么是JPA? 全称Java Persistence API,可以通过注解或者XML描述[对象-关系表]之间的映射关系,并将实体对象持久化到数据库中. 为我们提供了: 1)ORM映射元数据:JPA支 ...

  2. WCF SqlParameter序列化问题解决方案

    博文 http://www.cnblogs.com/pan11jing/archive/2011/08/19/2051827.html 通过自定义类,再在WCF端转换的方式解决问题,之后出现了一个很小 ...

  3. Java基础篇——集合浅谈

    原创不易,如需转载,请注明出处https://www.cnblogs.com/baixianlong/p/10703558.html,否则将追究法律责任!!! Set(基于Map来实现的,不细说) H ...

  4. virtualbox中 清理磁盘

    1. 碎片整理 windows: 下载 sdelete 工具 执行命令: sdelete –z c:\ Linux: 执行如下命令: sudo dd if=/dev/zero of=/EMPTY bs ...

  5. Https协议报错:com.sun.net.ssl.internal.www.protocol.https.HttpsURLConnectionOldImpl解决方法

    旭日Follow_24 的CSDN 博客 ,全文地址请点击: https://blog.csdn.net/xuri24/article/details/82220333 所用应用服务器:JBoss服务 ...

  6. C# % 和 /

    /相当于整数除法中的除号,%相当于余号5 除以 2 = 2 余 1,因此 5/2=2,5%2=1.

  7. K8S 部署 ingress-nginx (一) 原理及搭建

    Kubernetes 暴露服务的有三种方式,分别为 LoadBlancer Service.NodePort Service.Ingress.官网对 Ingress 的定义为管理对外服务到集群内服务之 ...

  8. python中经典类和新式类的区别

    要知道经典类和新式类的区别,首先要掌握类的继承.类的继承的一个优点就是减少代码,而且使代码看起来结构很完整. 那什么是经典类,什么是新式类呢? 经典类和新式类的主要区别就是类的继承的方式 ,经典类遵循 ...

  9. 2018-09-24 Java源码英翻中网页演示

    在线演示地址: 源代码翻译 两部分如下. 独立的Java代码翻译库 续前文代码翻译尝试-使用Roaster解析和生成Java源码 源码库: program-in-chinese/java_code_t ...

  10. loadrunner 脚本录制-录制选项设置HTML-based URL-based Script

    脚本录制-录制选项设置, HTML-based Script与URL-based Script by:授客 QQ:1033553122 Access:Vugen->Tool->Record ...