Portal

Portal1: Codeforces

Portal2: Luogu

Description

The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "\(\texttt{name score}\)", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to \(m\)) at the end of the game, than wins the one of them who scored at least \(m\) points first. Initially each player has \(0\) points. It's guaranteed that at the end of the game at least one player has a positive number of points.

Input

The first line contains an integer number \(n(1 \le n \le 1000)\), \(n\) is the number of rounds played. Then follow \(n\) lines, containing the information about the rounds in "\(\texttt{name score}\)" format in chronological order, where \(\texttt{name}\) is a string of lower-case Latin letters with the length from \(1\) to \(32\), and \(\texttt{score}\) is an integer number between \(-1000\) and \(1000\), inclusive.

Output

Print the name of the winner.

Sample Input1

3
mike 3
andrew 5
mike 2

Sample Output1

andrew

Sample Input2

3
andrew 3
andrew 2
mike 5

Sample Output2

andrew

Description in Chinese

Berland流行着纸牌游戏“Berlogging”,这个游戏的赢家是根据以下规则确定的:在每一轮中,玩家获得或失去一定数量的分数,在游戏过程中,分数被记录在“名称和得分”行中,其中名字是玩家的名字,得分是在这一轮中获得的分数。得分是负值意味着玩家失去了相应的分数。如果在比赛结束时只有一名玩家分数最多,他就是获胜者。如果两名或两名以上的玩家在比赛结束时都有最大的分数\(m\),那么其中首先获得至少\(m\)分的玩家胜利。开始时,每个玩家都是\(0\)分。保证在比赛结束时至少有一个玩家的分数为正。

Solution

把每个人的得分存到\(\texttt{map}\)里,然后求出最大得分,再寻找最先到达最大得分的人,即为获胜者。

Code

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map> using namespace std; const int INF = 0x3f3f3f3f, MAXN = 1005;
int n, a[MAXN];
string name[MAXN];
map<string, int> Map1, Map2;
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
cin >> name[i] >> a[i];
Map1[name[i]] += a[i];
}//映射到map里
int Max = -INF;
for (int i = 0; i < n; i++)
if (Map1[name[i]] > Max) Max = Map1[name[i]];//求出最大得分
for (int i = 0; i < n; i++) {
Map2[name[i]] += a[i];
if (Map2[name[i]] >= Max && Map1[name[i]] >= Max) {
cout << name[i] << endl;//输出最先到最大得分的人
return 0;
}
}
return 0;
}

『题解』Codeforces2A Winner的更多相关文章

  1. 『题解』洛谷P1063 能量项链

    原文地址 Problem Portal Portal1:Luogu Portal2:LibreOJ Portal3:Vijos Description 在\(Mars\)星球上,每个\(Mars\)人 ...

  2. 『题解』Codeforces1142A The Beatles

    更好的阅读体验 Portal Portal1: Codeforces Portal2: Luogu Description Recently a Golden Circle of Beetlovers ...

  3. 『题解』Codeforces1142B Lynyrd Skynyrd

    更好的阅读体验 Portal Portal1: Codeforces Portal2: Luogu Description Recently Lynyrd and Skynyrd went to a ...

  4. 『题解』洛谷P1993 小K的农场

    更好的阅读体验 Portal Portal1: Luogu Description 小\(K\)在\(\mathrm MC\)里面建立很多很多的农场,总共\(n\)个,以至于他自己都忘记了每个农场中种 ...

  5. 『题解』洛谷P2296 寻找道路

    更好的阅读体验 Portal Portal1: Luogu Portal2: LibreOJ Description 在有向图\(\mathrm G\)中,每条边的长度均为\(1\),现给定起点和终点 ...

  6. 『题解』洛谷P1351 联合权值

    更好的阅读体验 Portal Portal1: Luogu Portal2: LibreOJ Description 无向连通图\(\mathrm G\)有\(n\)个点,\(n - 1\)条边.点从 ...

  7. 『题解』Codeforces656E Out of Controls

    更好的阅读体验 Portal Portal1: Codeforces Portal2: Luogu Description You are given a complete undirected gr ...

  8. 『题解』洛谷P2170 选学霸

    更好的阅读体验 Portal Portal1: Luogu Description 老师想从\(N\)名学生中选\(M\)人当学霸,但有\(K\)对人实力相当,如果实力相当的人中,一部分被选上,另一部 ...

  9. 『题解』洛谷P1083 借教室

    更好的阅读体验 Portal Portal1: Luogu Portal2: LibreOJ Portal3: Vijos Description 在大学期间,经常需要租借教室.大到院系举办活动,小到 ...

随机推荐

  1. B-经济学-基尼指数

    目录 基尼指数 一.基尼指数简介 更新.更全的<机器学习>的更新网站,更有python.go.数据结构与算法.爬虫.人工智能教学等着你:https://www.cnblogs.com/ni ...

  2. MySQL基础(三)多表查询(各种join连接详解)

    Mysql 多表查询详解 一.前言 二.示例 三.注意事项 一.前言 上篇讲到Mysql中关键字执行的顺序,只涉及了一张表:实际应用大部分情况下,查询语句都会涉及到多张表格 : 1.1 多表连接有哪些 ...

  3. 域渗透-Kerberos协议中spn的应用

    0x01 关于SPN 服务主体名称(SPN)是Kerberos客户端用于唯一标识给特定Kerberos目标计算机的服务实例名称. 服务主体名称是服务实例(可以理解为一个服务,比如 HTTP.MSSQL ...

  4. [BZOJ4947] 字符串大师 - KMP

    4974: [Lydsy1708月赛]字符串大师 Time Limit: 1 Sec  Memory Limit: 256 MBSubmit: 739  Solved: 358[Submit][Sta ...

  5. 数据结构4_java---顺序串,字符串匹配算法(BF算法,KMP算法)

    1.顺序串 实现的操作有: 构造串 判断空串 返回串的长度 返回位序号为i的字符 将串的长度扩充为newCapacity 返回从begin到end-1的子串 在第i个字符之前插入字串str 删除子串 ...

  6. 算法问题实战策略 DICTIONARY

    地址 https://algospot.com/judge/problem/read/DICTIONARY 解法 构造一个26字母的有向图 判断无回路后 就可以输出判断出来的字符序了 比较各个字母的先 ...

  7. Hystrix dashboard - Unable to connect to Command Metric Stream.

    在使用boot 2.0.*以上版本 + cloud Finchley.RELEASE 查看仪表盘的时候会报错 Unable to connect to Command Metric Stream &l ...

  8. IIS服务器文件跨域问题(几乎可以解决大多数跨域问题)

    Windows server 服务器 1:在管理工具中选择,找到IIS 2:打开IIS管理,找到网站 3:找到HTTP响应头标 4:打开HTTP响应头标 5:添加 添加:Access-Control- ...

  9. JVM概述和类加载器

    JVM概述 1.Java虚拟机所管理的内存包括以下几个运行时数据区域:   ①.程序计数器     程序计数器是一块较小的内存空间,可以看做是当前线程所执行的字节码的行号指示器,字节码解释器工作时就是 ...

  10. 封装自己通用的 增删改查的方法 By EF

    封装自己的通用CURD By EF using System; using System.Collections.Generic; using System.Data.Entity; using Sy ...