The gaming company Sandstorm is developing an online two player game. You have been asked to implement the ranking system. All players have a rank determining their playing strength which gets updated after every game played. There are 25 regular ranks, and an extra rank, “Legend”, above that. The ranks are numbered in decreas- ing order, 25 being the lowest rank, 1 the second highest rank, and Legend the highest rank.

Each rank has a certain number of “stars” that one needs to gain before advancing to the next rank. If a player wins a game, she gains a star. If before the game the player was on rank 6-25, and this was the third or more consecutive win, she gains an additional bonus star for that win. When she has all the stars for her rank (see list below) and gains another star, she will instead gain one rank and have one star on the new rank.

For instance, if before a winning game the player had all the stars on her current rank, she will after the game have gained one rank and have 1 or 2 stars (depending on whether she got a bonus star) on the new rank. If on the other hand she had all stars except one on a rank, and won a game that also gave her a bonus star, she would gain one rank and have 1 star on the new rank. If a player on rank 1-20 loses a game, she loses a star. If a player has zero stars on a rank and loses a star, she will lose a rank and have all stars minus one on the rank below. However, one can never drop below rank 20 (losing a game at rank 20 with no stars will have no effect).

If a player reaches the Legend rank, she will stay legend no matter how many losses she incurs afterwards. The number of stars on each rank are as follows:

• Rank 25-21: 2 stars

• Rank 20-16: 3 stars

• Rank 15-11: 4 stars

• Rank 10-1: 5 stars

A player starts at rank 25 with no stars. Given the match history of a player, what is her rank at the end of the sequence of matches?

Input

There will be several test cases. Each case consists of a single line describing the sequence of matches. Each character corre- sponds to one game; ‘W’ represents a win and ‘L’ a loss. The length of the line is between 1 and 10 000 characters (inclusive).

Output

Output a single line containing a rank after having played the given sequence of games; either an integer between 1 and 25 or “Legend”.

Sample Input

WW
WWW
WWWW
WLWLWLWL
WWWWWWWWWLLWW
WWWWWWWWWLWWL

Sample Output

25
24
23
24
19
18

Hint

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main ()
{
char s[11000];
while(~scanf("%s",s))
{
int len = strlen(s);
int star = 0;
int win = 0; int rank = 25;
for (int i = 0; i < len; i++)
{
if (rank == 0)
{
break;
}
if (s[i] == 'W')
{
win ++;
star++; if (win >= 3 && rank > 5)
star++; if (rank<=25 && rank>=21 && star > 2)
{
rank--;
star-=2;
}
if (rank>=16 && rank<=20 && star > 3)
{
rank--;
star-=3;
}
if (rank>=11 && rank<=15 && star > 4)
{
rank--;
star-=4;
}
if (rank>=1 && rank<=10 && star > 5)
{
rank--;
star-=5;
}
}
else
{
win = 0;
if (rank>=21)
continue;
star --;
if (star < 0)
{
if (rank == 20)
{
star = 0;
}
if (rank>=15 && rank<=19)
{
rank++;
star = 2;
}
if (rank>=10 && rank<15)
{
rank++;
star = 3;
} if (rank>=1 && rank<=9)
{
rank++;
star = 4;
}
}
}
} if (!rank)
cout<<"Legend"<<endl;
else
cout<<rank<<endl;
} return 0; }

CSU-2018的更多相关文章

  1. CSU 2018年12月月赛 H(2220): Godsend

    Description Leha somehow found an array consisting of n integers. Looking at it, he came up with a t ...

  2. CSU 2018年12月月赛 G(2219): Coin

    Description 有这样一个众所周知的问题: 你面前有7个硬币,其中有一个劣质的(它比正常的硬币轻一点点),你有一个天平,问需要你需要使用天平多少次能保证找到那个劣质的硬币. 众所周知的算法是: ...

  3. CSU 2018年12月月赛 F(2218): Finding prime numbers

    Description xrdog has a number set. There are 95 numbers in this set. They all have something in com ...

  4. CSU 2018年12月月赛 D 2216 : Words Transformation

    Description There are n words, you have to turn them into plural form. If a singular noun ends with ...

  5. CSU 2018年12月月赛 B 2214: Sequence Magic

    Description 有一个1到N的自然数序列1,2,3,...,N-1,N. 我们对它进行M次操作,每次操作将其中连续的一段区间 [Ai,Bi][Ai,Bi] (即第Ai个元素到第Bi个元素之间的 ...

  6. CSU 2018年12月月赛 A 2213: Physics Exam

    Description 高中物理老师总认为给学生文本形式的问题比给纯计算形式的问题要求更高.毕竟,学生首先得阅读和理解问题. 因此,他们描述一个问题不像”U=10V,I=5A,P=?”,而是”有一个含 ...

  7. 2018. The Debut Album

    http://acm.timus.ru/problem.aspx?space=1&num=2018 真心爱过,怎么能彻底忘掉 题目大意: 长度为n的串,由1和2组成,连续的1不能超过a个,连续 ...

  8. Math.abs(~2018),掌握规律即可!

    Math.abs(~2018) 某前端群的入门问题长姿势了,一个简单的入门问题却引发了我的思考,深深的体会到自己在学习前端技术的同时忽略遗忘了一些计算机的基础知识. 对于 JS Math对象没什么可说 ...

  9. csu 1812: 三角形和矩形 凸包

    传送门:csu 1812: 三角形和矩形 思路:首先,求出三角形的在矩形区域的顶点,矩形在三角形区域的顶点.然后求出所有的交点.这些点构成一个凸包,求凸包面积就OK了. /************** ...

  10. CSU 1503 点到圆弧的距离(2014湖南省程序设计竞赛A题)

    题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1503 解题报告:分两种情况就可以了,第一种是那个点跟圆心的连线在那段扇形的圆弧范围内,这 ...

随机推荐

  1. K8S入门系列之集群二进制部署-->master篇(二)

    组件版本和配置策略 组件版本 Kubernetes 1.16.2 Docker 19.03-ce Etcd 3.3.17 https://github.com/etcd-io/etcd/release ...

  2. Dubbo的应用

    导语:Dubbo是阿里巴巴的一个分布式服务的开源框架,致力于提供高性能和透明化的RPC远程服务调用方案,是阿里巴巴SOA服务化治理方案的核心框架,每天为2,000+个服务提供3,000,000,000 ...

  3. 安装&卸载Windows服务

    使用.NET Framework的工具InstallUtil.exe. 安装服务 C:\Windows\Microsoft.NET\Framework\v4.0.30319\InstallUtil.e ...

  4. T-SQL Part IV: ORDER BY

    ORDER BY 返回一个Cursor,并不返回结果集.而试图将Cursor作为输入将产生了错误. 所以,下列的SQL语句将产生错误: SELECT VerID, IsComplete VerID, ...

  5. MySQL InnoDB MVCC

    MySQL 原理篇 MySQL 索引机制 MySQL 体系结构及存储引擎 MySQL 语句执行过程详解 MySQL 执行计划详解 MySQL InnoDB 缓冲池 MySQL InnoDB 事务 My ...

  6. SpringBoot学习(二)—— springboot快速整合spring security组件

    Spring Security 简介 spring security的核心功能为认证(Authentication),授权(Authorization),即认证用户是否能访问该系统,和授权用户可以在系 ...

  7. 稀疏数组 python描述

    什么是稀疏矩阵? 在矩阵中,若数值为0的元素数目远远多于非0元素的数目,并且非0元素分布没有规律时,则称该矩阵为稀疏矩阵. 作用: 在这种情况下,很多0值无疑是很浪费空间的,当我们要把数组存储在磁盘中 ...

  8. 2019-9-16:渗透测试,基础学习,Linux下软件安装,环境搭建,笔记

    Centos linux下软件安装yum 通过分析rpm包头数据后,自动解决依赖关系,直接云端下载软件,根据不同版本系统获取不同软件信息,按顺序下载rpm包,安装软件yum search 软件名:搜索 ...

  9. (四)OpenStack---M版---双节点搭建---Glance安装和配置

    ↓↓↓↓↓↓↓↓视频已上线B站↓↓↓↓↓↓↓↓ >>>>>>传送门 1.创建glance数据库 2.获得 admin 凭证来获取只有管理员能执行的命令的访问权限 3 ...

  10. 通过 Python 理解 Mixin 概念

    Mixin 的概念 Mixin 即 Mix-in,常被译为"混入",是一种编程模式,在 Python 等面向对象语言中,通常它是实现了某种功能单元的类,用于被其他子类继承,将功能组 ...