codeforces-3A-A. Winner （C++ && 简单模拟）

65 阅读 0 评论 43 点赞

我是靠谱客的博主潇洒香烟，这篇文章主要介绍codeforces-3A-A. Winner （C++ && 简单模拟），现在分享给大家，希望可以做个参考。

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.

        Examples 
      

          input 
        
3
mike 3
andrew 5
mike 2

          output 
        
andrew

          input 
        
3
andrew 3
andrew 2
mike 5

          output 
        
andrew题目意思就是给出一个各轮选手得分,最后输出分最高的那个人.
如果有多人分数一样高,那就输出先达到这个分数的人.
简单模拟,用stl的map来做.
1.统计每人分数,顺便把每轮的情况也记下来.
2.统计出最高分.
3.遍历每个人的总得分,找到分数最高的那些人.
4.遍历每轮的情况,第一个达到最高分的人,输出就好了.
#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，这里的大于号就是考虑到出现负数的情况，分数越来越小
            {  
                cout<<name[i]<<endl;  
                return 0;  
            }  
        }  
    }  

    return 0;  
}