CF750C New Year and Rating

题意: 在cf系统中,给定一个人每次比赛所属的div及比赛之后的分数变化.计算经过这些比赛之后此人的rating最高可能是多少. 思路: 模拟. 实现: #include <cstdio> #include <iostream> #include <algorithm> using namespace std; const int INF = 0x3f3f3f3f; int n, minn = -INF, maxn = INF, c, x; int main() { s…

Codefroces 750C：New Year and Rating（思维）

http://codeforces.com/contest/750/problem/C 题意:有n场比赛,每场比赛有一个c,代表比赛结束后分数的增长情况,有一个d,代表这场比赛在div1或者div2打.div1 >= 1900,div2 < 1900.求打完这n场比赛后可能得到的最大分数.如果无限大就输出"Inifinity",如果有矛盾就输出"Impossible". 思路:官方题解. For every contest we know that th…

AngularJs的UI组件ui-Bootstrap分享（十二）——Rating

Rating是一个用于打分或排名的控件.看一个最简单的例子: <!DOCTYPE html> <html ng-app="ui.bootstrap.demo" xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /&…

从Elo Rating System谈到层次分析法

1. Elo Rating System Elo Rating System对于很多人来说比较陌生,根据wikipedia上的解释:Elo评分系统是一种用于计算对抗比赛(例如象棋对弈)中对手双方技能水平的方法,它由Arpad Elo创建.(The Elo rating system is a method for calculating the relative skill levels of players in competitor-versus-competitor games such…

HDU 4870 Rating（概率、期望、推公式） && ZOJ 3415 Zhou Yu

其实zoj 3415不是应该叫Yu Zhou吗...碰到ZOJ 3415之后用了第二个参考网址的方法去求通项,然后这次碰到4870不会搞.参考了chanme的,然后重新把周瑜跟排名都反复推导(不是推倒)四五次才上来写这份有抄袭嫌疑的题解... 这2题很类似,多校的rating相当于强化版,不过原理都一样.好像是可以用高斯消元做,但我不会.默默推公式了. 公式推导参考http://www.cnblogs.com/chanme/p/3861766.html#2993306 http://www.cn…

Elo rating system 模拟

package org.cc.foo_008; import java.util.ArrayList; import java.util.List; import java.util.Random; public class Main_006 { public static void main(String[] args) { EloRatingSystemDemo e=new EloRatingSystemDemo(); e.show(); // User u=e.gamer.get(new…

HDU4870 Rating(概率)

第一场多校,感觉自己都跳去看坑自己的题目里去了,很多自己可能会比较擅长一点的题目没看,然后写一下其中一道概率题的题解吧,感觉和自己前几天做的概率dp的思路是一样的.下面先来看题意:一个人有两个TC的账号,一开始两个账号rating都是0,然后每次它会选择里面rating较小的一个账号去打比赛,每次比赛有p的概率+1分,有1-p的概率-2分,当然如果本身是<=2分的也就还是回到0分.然后问最后其中一个账号到达20分时需要打多少次比赛. 先考虑一场比赛的情况,定义dp[k]为当前为k分,要达到20分…

HDU 4870 Rating 概率DP

Rating Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4870 Description A little girl loves programming competition very much. Recently, she has found a new kind of programming competition named…

hdu 4870 Rating

题目链接:hdu 4870 这题应该算是概率 dp 吧,刚开始看了好几个博客都一头雾水,总有些细节理不清楚,后来看了 hdu 4870 Rating (概率dp) 这篇博客终于有如醍醐灌顶,就好像是第 7 个烧饼一样,终于感到饱了. 还是适当贴下图先: 按照这个思路我的代码如下: #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ]; int main() { doub…

HDU 4870 Rating（高斯消元）

HDU 4870 Rating 这是前几天多校的题目,高了好久突然听旁边的大神推出来说是可以用高斯消元,一直喊着赶快敲模板,对于从来没有接触过高斯消元的我来说根本就是一头雾水,无赖之下这几天做DP,正好又做到了这个题,没办法得从头开始看,后来在网上找了别人的高斯消元的模板后发现其实也还是很好理解,就是先构造一个增广矩阵,然后化行阶梯形,最后迭代求解首先有一个介绍高斯消元感觉过于详细的博客http://blog.csdn.net/tsaid/article/details/7329301 首…

2014多校第一场J题 || HDU 4870 Rating（DP || 高斯消元）

题目链接题意 :小女孩注册了两个比赛的帐号,初始分值都为0,每做一次比赛如果排名在前两百名,rating涨50,否则降100,告诉你她每次比赛在前两百名的概率p,如果她每次做题都用两个账号中分数低的那个去做,问她最终有一个账号达到1000分需要做的比赛的次数的期望值. 思路 :可以直接用公式推出来用DP做,也可以列出210个方程组用高斯消元去做. (1)DP1:离散化.因为50,100,1000都是50的倍数,所以就看作1,2,20.这样做起来比较方便. 定义dp[i]为从 i 分数到达i+1…

HDU 4870 Rating (2014 Multi-University Training Contest 1)

Rating Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 578 Accepted Submission(s): 363Special Judge Problem Description A little girl loves programming competition very much. Recently, she h…

Rating

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 786 Accepted Submission(s): 496Special Judge Problem Description A little girl loves programming competition very much. Recently, she has fou…

HDU 4870 Rating (2014 多校联合第一场 J)(概率)

题意: 一个人有两个TC的账号,一开始两个账号rating都是0,然后每次它会选择里面rating较小的一个账号去打比赛,每次比赛有p的概率+1分,有1-p的概率-2分,当然如果本身是<=2分的也就还是回到0分.然后问最后其中一个账号到达20分时需要打多少次比赛. 思路: 因为每次50分,到达1000分,所以可以看做每次1分,到达20分dp[i]表示i到20的数学期望那么dp[i] = dp[i+1]*p+dp[i-2]*q+1;令t[i] = dp[i+1]-dp[i]则t[i] = (t[i…

2014 Multi-University Training Contest 1 - J Rating

题目链接: pid=4870">http://acm.hdu.edu.cn/showproblem.php?pid=4870 题目大意: 题意:一个人注冊两个账号,初始rating都是0,他每次拿低分的那个号去打比赛,赢了加50分,输了扣100分,胜率为p,他会打到直到一个号有1000分为止,问比赛场次的期望. 解题思路: 用E(x.y)表示到(1000, 950)这个状态须要多少场.由于(1000,X) X != 950 这样的情况是不可能发生的.最后E(0,0)就是答案. E(x1,…

HDU4870:Rating(DP)

Problem Description A little girl loves programming competition very much. Recently, she has found a new kind of programming competition named "TopTopTopCoder". Every user who has registered in "TopTopTopCoder" system will have a ratin…

hdu 4870 Rating(可能性DP&高数消除)

Rating Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 714 Accepted Submission(s): 452 Special Judge Problem Description A little girl loves programming competition very much. Recently, she…

C. New Year and Rating

C. New Year and Rating time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Every Codeforces user has rating, described with one integer, possibly negative or zero. Users are divided into two d…

2018年多校第三场第一题 A. Ascending Rating hdu6319

比赛地址:http://acm.hdu.edu.cn/contests/contest_show.php?cid=804 题目编号:第一题 A. Ascending Rating hdu6319 题意: 给定一个序列 a[1..n],对于每个长度为 m 的连续子区间, 求出区间 a 的最大值以及从左往右扫描该区间时 a 的最大值的变化次数. 题解:按照 r 从 m 到 n 的顺序很难解决这个问题. 考虑按照 r 从 n 到 m 的顺序倒着求出每个区间的答案. 按照滑窗最大值的经典方法维护 a…

Spark 中Java实现数据库Row转Rating

Dataset<Row> ratings = mlsc.sql("SELECT user,movie,rating FROM data");JavaRDD<Row> rowJavaRDD = ratings.javaRDD(); JavaRDD<String[]> stringJavaRDD = rowJavaRDD.map(new Function<Row, String[]>() { @Override public String[]…

hdu多校第3场A.Ascending Rating

Problem A. Ascending Rating Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Before the start of contest, there are n ICPC contestants waiting to n from left to right. It…

杭电多校第三场 A Ascending Rating

Problem Description Before the start of contest, there are n ICPC contestants waiting in a long queue. They are labeled by 1 to n from left to right. It can be easily found that the i-th contestant's QodeForces rating is ai.Little Q, the coach of Qua…

Ascending Rating（单调队列）

题目描述 Before the start of contest, there are n ICPC contestants waiting in a long queue. They are labeled by 1 to n from left to right. It can be easily found that the i-th contestant's QodeForces rating is ai. Little Q, the coach of Quailty Normal Un…

SharePoint中Rating相关的字段。

From: https://sharepoint.stackexchange.com/questions/194197/how-to-manipulate-likeby-nooflikes-ratedby-averagerating-userrating-using-cs Copy below: Gone through the problem an got the relevant answer also made blog on it so visit the blog post S…

jQuery生成一个DIV容器，ID是"rating".

我们需要一些服务器端代码,这个例子中用到了一个PHP文件,读取rating参数然后返回rating总数和平均数.看一下rate.php代码.虽然这些例子也可以不使用AJAX来实现,但显示我们不会那么做,我们用jQuery生成一个DIV容器,ID是"rating". $(document).ready(function() { // generate markup var ratingMarkup = ["lease rate: "]; …

HDU 2018 Multi-University Training Contest 3 Problem A. Ascending Rating 【单调队列优化】

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=6319 Problem A. Ascending Rating Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 5943 Accepted Submission(s): 2004 Problem Description Before…

POJ1030 Rating

题目来源:http://poj.org/problem?id=1030 题目大意:有100支队伍(编号1->100),有两场比赛.以下表的形式列出了两场比赛的名次.(有的队伍没有参赛或只参加了一场比赛.) 要求两场比赛的总排名.计算规则是: 1.如果某队两场都比另一队排名靠前(或一场赢一场平),则总排名靠前. 2.如果两队在两场比赛中各赢一场,则他们的排名取决于他们在两场比赛中的排名差.如上面的例子中1队在比赛1中赢了5队,名次差为3,在比赛2中输给5对,名次差为1,所以1队应排在5队前面.如果…