• Codeforces 2A :winner


    A. Winner
    time limit per test
    1 second
    memory limit per test
    64 megabytes
    input
    standard input
    output
    standard output

    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 "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  ≤  n  ≤  1000), n is the number of rounds played. Then follow n lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.

    Output

    Print the name of the winner.

    Sample test(s)
    input
    3
    mike 3
    andrew 5
    mike 2
    
    output
    andrew
    
    input
    3
    andrew 3
    andrew 2
    mike 5
    
    output
    andrew
    

    做Codeforces上的题还很少,但只从这道题来说,这题无论从题意还是各个方面都足够有趣。一共有n轮比赛,接下来的每行代表这一轮比赛中,这个人得到的分数,分数可正可负。问最先到达最大分值的人是谁。

    大一的时候做这个居然用的是数组,现在用map果然方便太多了,但还是因为一开始没注意有可能是负数导致,判断的时候出现了问题。

    一共判断的时候有两轮,第一轮判断这个人的结果是不是最大值,如果是最大值,就要看它是不是第一个到达的,也即第二轮的成绩是不是大于或是等于最大值,如果满足条件,赶紧输出结果,就行了。最近做题被虐的都没人样了。。。


    代码:

    #include <iostream>
    #include <map>
    #include <cstring>
    #include <string>
    #include <algorithm>
    using namespace std;
    
    //一开始没有考虑到负数的情况,出现了错误。出现负数的话,分数就会下降,之前的最大值就不会对。
    
    map<string,int> round;
    map<string,int> round2;
    
    string name[1005];
    int score[1005];
    
    int main()
    {
    	int count,i;
    	cin>>count;
    
    	round.clear();
    
    	int max = -1000005;
    	string max_name;
    
    	for(i=1;i<=count;i++)
    	{
    		cin>>name[i]>>score[i];
    
    		round[name[i]] += score[i];
    	}
    
    	for(i=1;i<=count;i++)
    	{
    		if(round[name[i]]>max)
    		{
    			max=round[name[i]];
    		}
    	}
    	
    	for(i=1;i<=count;i++)
    	{
    		if(round[name[i]] == max)
    		{
    			round2[name[i]] += score[i];
    
    			if(round2[name[i]]>=max)//最先到达
    			{
    				cout<<name[i]<<endl;
    				return 0;
    			}
    		}
    	}
    
    	return 0;
    }
    


    版权声明:本文为博主原创文章,未经博主允许不得转载。

  • 相关阅读:
    游戏编程模式之事件队列模式
    游戏编程模式之组件模式
    游戏编程模式之类型对象模式
    游戏编程模式之父类沙盒模式
    游戏编程模式之字节码模式
    游戏人工智能简介
    游戏编程模式之更新模式
    游戏编程模式之游戏循环
    .vimrc配置文件 + gvim 运行 gnome-terminal 完成后等待
    js 批量移除steam游戏 移除用户凭证中免费获取的物品
  • 原文地址:https://www.cnblogs.com/lightspeedsmallson/p/4785883.html
Copyright © 2020-2023  润新知