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
- 2019-04-17 PowerShell基本语法
打印Hello World ,Hello theDataDigger writeLog ' Hello World'$Name = "theDataDigger"writeLog ...
- 数据库工具——Navicat Premium使用技巧
Navicat Premium 常用功能讲解 1.快捷键 1.1. F8 快速回到当前对象列表 1.2. Ctrl + q 打开查询界面 1.3. Ctrl + d 快速修改当前的表结构 1.4 ...
- 洛谷 P2728 纺车的轮子 Spinning Wheels
P2728 纺车的轮子 Spinning Wheels 题目背景 一架纺车有五个纺轮(也就是五个同心圆),这五个不透明的轮子边缘上都有一些缺口.这些缺口必须被迅速而准确地排列好.每个轮子都有一个起始标 ...
- Shell、Xterm、Gnome-Terminal、Konsole简介(转)
什么是Shell? 简单的说, Shell就是一个小程序,这个小程序可以接受来自键盘的命令并把这些命令发送到操作系统,再有系统来执行.在过去,在安装有Unix的计算机上,这是唯一的可用的交互式操作.而 ...
- HDU 1257(最小拦截系统)
Description 某国为了防御敌国的导弹袭击,发展出一种导弹拦截系统.但是这种导弹拦截系统有一个缺陷:虽然它的第一发炮弹能够到达任意的高度,但是以后每一发炮弹都不 能超过前一发的高度.某天,雷达 ...
- 用虚拟机创建win7 32位系统来测试win 7 64位系统无法安装cad 2004 缺少acdb16.dll的问题
- 43.qt通过qss自定义外观
样式: 文件格式类型: candy.qss /* R1 */ QDialog { /*设置背景图片*/ background-image: url(:/images/background.png); ...
- B - Guess a number!
Problem description A TV show called "Guess a number!" is gathering popularity. The whole ...
- 带中横线的日期格式在iOS手机系统上 转换时间戳NaN问题
类似于 '2019-04-01 14:13:00' 这样的日期格式转换时间戳在iOS手机上是无法转换的,需要先处理日期格式成 '2019/04/01 14:13:00' var str = '2019 ...
- (转)(C++)关于抽象基类和纯虚函数
★抽象类:一个类可以抽象出不同的对象来表达一个抽象的概念和通用的接口,这个类不能实例化(创造)对象. ★纯虚函数(pure virtual):在本类里不能有实现(描述功能),实现需要在子类中实现.例: ...