Kattis - Game Rank
Game Rank
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 2525 regular ranks, and an extra rank, “Legend”, above that. The ranks are numbered in decreasing order, 2525 being the lowest rank, 11 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 66-2525, 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 11 or 22 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 11 star on the new rank.
If a player on rank 11-2020 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 2020(losing a game at rank 2020 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 2525-2121: 22 stars
Rank 2020-1616: 33 stars
Rank 1515-1111: 44 stars
Rank 1010-11: 55 stars
A player starts at rank 2525 with no stars. Given the match history of a player, what is her rank at the end of the sequence of matches?
Input
The input consists of a single line describing the sequence of matches. Each character corresponds to one game; ‘W’ represents a win and ‘L’ a loss. The length of the line is between 11 and 1000010000 characters (inclusive).
Output
Output a single line containing a rank after having played the given sequence of games; either an integer between 11 and 2525 or “Legend”.
| Sample Input 1 | Sample Output 1 |
|---|---|
WW |
25 |
| Sample Input 2 | Sample Output 2 |
|---|---|
WWW |
24 |
| Sample Input 3 | Sample Output 3 |
|---|---|
WWWW |
23 |
| Sample Input 4 | Sample Output 4 |
|---|---|
WLWLWLWL |
24 |
| Sample Input 5 | Sample Output 5 |
|---|---|
WWWWWWWWWLLWW |
19 |
| Sample Input 6 | Sample Output 6 |
|---|---|
WWWWWWWWWLWWL |
18 |
题意
25级升24级,要两颗星,但是不是两颗星满了升24,是3颗星才升级变成24级一星。然后24到23到……一直到20都是两颗星,然后3颗星 4颗星 5颗星,然后你20级以下是不掉级的,输了也不掉,如果你是20级0星不会掉星,但是20级一颗星就会掉,你掉级条件是,当前星数为0而且输了,那就掉级了,然后,他还有一个设定,如果在5级以下连胜三局或者以上,一局奖励两颗星,然后,一级是顶级,超过了一级就直接输出LEGEND,接下来那个人输成什么样都是LEGEND,其实就是炉石传说的规则
代码
#include<bits/stdc++.h>
using namespace std;
int k[] = {, , , , , , , , , , , , , , , , , , , , , , , , , };
char aa[];
int main() {
int n;
while (~scanf("%s", aa)) {
n = strlen(aa);
int ans = , b = ;
for (int i = ; i < n; i++) {
if (aa[i] == 'W') {
b++;
if (i > && aa[i - ] == 'W' && aa[i - ] == 'W' && ans >= )b++;
if (b > k[ans]) {
b -= k[ans]; ans--;
}
} else {
if (ans > || ans == && b == ) {
continue;
}
b--;
if (b < ) {
ans++; b = k[ans] - ;
}
}
if (ans == )break;
}
if (ans) {
printf("%d\n", ans);
} else {
puts("Legend");
}
}
return ;
}
Kattis - Game Rank的更多相关文章
- UVA, 10336 Rank the Languages
难点在于:递归函数和输出: #include <iostream> #include <vector> #include <algorithm> #include ...
- [LeetCode] Rank Scores 分数排行
Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ra ...
- rank()函数的使用
排序: ---rank()over(order by 列名 排序)的结果是不连续的,如果有4个人,其中有3个是并列第1名,那么最后的排序结果结果如:1 1 1 4select scoreid, stu ...
- [转]oracle分析函数Rank, Dense_rank, row_number
oracle分析函数Rank, Dense_rank, row_number 分析函数2(Rank, Dense_rank, row_number) 目录 ==================== ...
- 分区函数Partition By的与row_number()的用法以及与排序rank()的用法详解(获取分组(分区)中前几条记录)
partition by关键字是分析性函数的一部分,它和聚合函数不同的地方在于它能返回一个分组中的多条记录,而聚合函数一般只有一条反映统计值的记录,partition by用于给结果集分组,如果没有指 ...
- Learning to rank 介绍
PS:文章主要转载自CSDN大神hguisu的文章"机器学习排序": http://blog.csdn.net/hguisu/article/details/79 ...
- R语言排序:sort(),rank(),order()示例
> x<-c(97,93,85,74,32,100,99,67) > sort(x) [1] 32 67 74 85 93 97 99 100 > order(x) [1] 5 ...
- [Machine Learning] Learning to rank算法简介
声明:以下内容根据潘的博客和crackcell's dustbin进行整理,尊重原著,向两位作者致谢! 1 现有的排序模型 排序(Ranking)一直是信息检索的核心研究问题,有大量的成熟的方法,主要 ...
- sqlserver 中row_number,rank,dense_rank,ntile排名函数的用法
1.row_number() 就是行号 2.rank:类似于row_number,不同之处在于,它会对order by 的字段进行处理,如果这个字段值相同,那么,行号保持不变 3.dense_rank ...
随机推荐
- Centos 修改主机名称
Centos 配置主机名称: 1.首先查询一下当前的主机名称 [root@localhost~]# hostnamectl status Static hostname: ****** //永久主机名 ...
- web开发如何使用高德地图API(一)浏览器定位
说两句: 以下内容除了我自己写的部分,其他部分在高德开放平台都有(可点击外链访问). 我所整理的内容以实际项目为基础希望更有针对性的,更精简. 点击直奔主题. 准备工作: 首先,注册开发者账号,成为高 ...
- PHP tools for Visual Studio 2013 安装、破解、配置教程
安装 首先,必须要安装vs2013.本人安装的是社区版,免费的同时功能又全面. 然后,去http://download.csdn.net/detail/liangzehong007/9076855 或 ...
- Timus - 1209 - 1, 10, 100, 1000...
先上题目: 1209. 1, 10, 100, 1000... Time limit: 1.0 secondMemory limit: 64 MB Let's consider an infinite ...
- javascript Prototype constructor的理解(转)
讲JS的构造的,这个比较清晰,但并不表示一定正确. 这几天一直在思考这个东东,感觉比以前理解更深入了. http://blog.csdn.net/chunqiuwei/article/details/ ...
- emacs 搭建racket开发环境
emacs 搭建racket开发环境 emacs下搭建开发racket的环境,笔者之前用过下面两种模式:geiser和racket-mode.相对而言,后一种方式要显得简单.本文主要介绍后一种方式环境 ...
- codeforces Round #258(div2) C解题报告
C. Predict Outcome of the Game time limit per test 2 seconds memory limit per test 256 megabytes inp ...
- UVA 10888 - Warehouse(二分图完美匹配)
UVA 10888 - Warehouse option=com_onlinejudge&Itemid=8&page=show_problem&category=562& ...
- oc16--set,get
// // Kline.h // day13 #import <Foundation/Foundation.h> @interface Kline : NSObject { int _ma ...
- Beta 分布归一化的证明(系数是怎么来的),期望和方差的计算
1. Γ(a+b)Γ(a)Γ(b):归一化系数 Beta(μ|a,b)=Γ(a+b)Γ(a)Γ(b)μa−1(1−μ)b−1 面对这样一个复杂的概率密度函数,我们不禁要问,Γ(a+b)Γ(a)Γ(b) ...