GYM 101550 G.Game Rank(模拟)
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
分析:
一个游戏级别分2525级和额外等级LegendLegend,其中2525级最低级,11级次高级,LegendLegend最高级,每个级别对应一个星星数量,对应表如下:
• Rank 25-21: 2 stars
• Rank 20-16: 3 stars
• Rank 15-11: 4 stars
• Rank 10-1: 5 stars
规则如下:
W:rk大于等于6小于等于25且3连胜的话,star加2,否则加一
stai满了之后,晋级:比如23级,2颗星再加上两颗星就是22,两颗星!
L:rk大于20的话,输了也不掉star
rk小于等于20,且star!=0的话,减少一颗星
rk<20且star==0的话,掉级,star数等于改级star总数-1
写了很久啊,妈卖批,规则很模糊
code:
#include<iostream>
using namespace std;
int rk,star,flag;
void down()
{
if(rk>)
return ;
if(star)
{
star--;
return ;
}
if(rk<)
{
rk++;
if(rk>=&&rk<=)
star=;
else if(rk>=&&rk<=)
star=;
else if(rk>=&&rk<=)
star=;
}
}
void up(int v)
{
star+=v;
if(rk>=&&rk<=&&star>)
{
rk--;
star-=;
}else if(rk>=&&rk<=&&star>)
{
rk--;
star-=;
}else if(rk>=&&rk<=&&star>)
{
rk--;
star-=;
}else if(rk>=&&rk<=&&star>)
{
rk--;
star-=;
}
}
void f(string str)
{
int l=str.length();
for(int i=;i<l;i++)
{
if(rk==)
break;
if(str[i]=='W')
{
flag++;
if(rk>=&&flag>=)
{
up();
}else
{
up();
}
}else
{
down();
flag=;
}
}
}
int main()
{
string str;
while(cin>>str)
{
rk=,star=,flag=;//初始化
f(str);
if(rk)
{
cout<<rk<<endl;
}else
{
cout<<"Legend"<<endl;
}
}
return ;
}
GYM 101550 G.Game Rank(模拟)的更多相关文章
- Codeforces Gym 100269B Ballot Analyzing Device 模拟题
Ballot Analyzing Device 题目连接: http://codeforces.com/gym/100269/attachments Description Election comm ...
- 牛客小白月赛2 G 文 【模拟】
链接:https://www.nowcoder.com/acm/contest/86/G来源:牛客网 题目描述 Sεlιнα(Selina) 开始了新一轮的男友海选.她要求她的男友要德智体美劳样样都全 ...
- Codeforces Gym 100513G G. FacePalm Accounting
G. FacePalm Accounting Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100513 ...
- Codeforces Gym 100637G G. #TheDress 暴力
G. #TheDress Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100637/problem/G ...
- Codeforces Gym 100513G G. FacePalm Accounting 暴力
G. FacePalm Accounting Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100513 ...
- Gym 100646 Problem C: LCR 模拟题
Problem C: LCR 题目连接: http://codeforces.com/gym/100646/attachments Description LCR is a simple game f ...
- Gym - 101147G G - The Galactic Olympics —— 组合数学 - 第二类斯特林数
题目链接:http://codeforces.com/gym/101147/problem/G G. The Galactic Olympics time limit per test 2.0 s m ...
- Gym 100952 G. The jar of divisors
http://codeforces.com/gym/100952/problem/G G. The jar of divisors time limit per test 2 seconds memo ...
- Codeforces Gym 100203G G - Good elements 标记暴力
G - Good elementsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/ ...
随机推荐
- 九度oj题目1385:重建二叉树
题目1385:重建二叉树 时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:4419 解决:1311 题目描述: 输入某二叉树的前序遍历和中序遍历的结果,请重建出该二叉树.假设输入的前序遍历和 ...
- 4.爬虫 requests库讲解 GET请求 POST请求 响应
requests库相比于urllib库更好用!!! 0.各种请求方式 import requests requests.post('http://httpbin.org/post') requests ...
- 微信小程序转百度小程序修改
百度小程序对比微信小程序(最初版):[设备]项里没有内存监控.iBeacon.wifi.蓝牙.用户截屏.手机联系人.NFC[位置]项里没有打开地图选择位置[界面]项里没有绘图功能.没有节点信息获取功能 ...
- 在浏览器中对访问的网页中的cookie添加和修改
做权限相关的东西,使用到了cookie,关于它的安全性,cookie在浏览器中,通过插件是可以对其进行修改的,如下: 1.FireFox 安装Edit This Cookie 插件,之后点击插件图标即 ...
- phpstorm一些简单配置
1.字体大小和行间距 2.设置编码:包括编辑工具编码和项目编码
- 微信小程序支付c#后台实现
今天为大家带来比较简单的支付后台处理 首先下载官方的c#模板(WxPayAPI),将模板(WxPayAPI)添加到服务器上,然后在WxPayAPI项目目录中添加两个“一般处理程序” (改名为GetOp ...
- bootstrap dialog对话框,完成操作提示框
1. 依赖文件: bootstrap.js bootstrap-dialog.js bootstrap.css bootstrap-dialog.css 2.代码 BootstrapDialog.co ...
- MySQL8.0加载文件内容报错: ERROR 1148: The used command is not allowed with this MySQL version
mysql数据库将文件内容加载到表中报错: mysql> LOAD DATA LOCAL INFILE '/path/pet.txt' INTO TABLE pet LINES TERMINAT ...
- 引爆你的Javascript代码进化 (转)
转自 海玉的博客 方才在程序里看到一段JS代码,写法极为高明,私心想着若是其按照规范来写,定可培养对这门语言的理解,对JS编程能力提高必是极好的.说人话:丫代码写的太乱,看的窝火! 最近闲暇无事,准备 ...
- JSTL标签概述
什么是JSTL JSP 标准标记库(JSP Standard Tag Library,JSTL)是一个实现 Web 应用程序中常见的通用功能的定制标记库集,这些功能包括迭代和条件判断.数据管理格式化. ...