zoukankan      html  css  js  c++  java
  • CodeForces 2A

    题目链接:http://codeforces.com/problemset/problem/2/A


    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

    题意:

    给出一些列的名字和分数!

    正的表示加分,负的表示减分! 求终于分数最大的人的名字;

    假设分数最大的人有多个,输出最先达到最大分数的人。

    代码例如以下:

    #include <cstdio>
    #include <iostream>
    #include <map>
    using namespace std;
    map<string, int> a,b;
    string s[1017];
    
    int main()
    {
        int x[1017];
        int n;
        cin >> n;
        for(int i = 1; i <= n; i++)
        {
            cin >> s[i] >> x[i];
            a[s[i]]+=x[i];
        }
        int maxx = 0;
        for(int i = 1; i <= n; i++)
        {
            if(a[s[i]] > maxx)
                maxx = a[s[i]];
        }
    
        for(int i = 1; i <= n; i++)
        {
            b[s[i]]+=x[i];
            if((b[s[i]]>=maxx) && (a[s[i]]>=maxx))//在终于分数是最大的人中,选首先达到最大分数的人
            {
                cout << s[i];
                break;
            }
        }
        return 0;
    }
    


  • 相关阅读:
    ubuntu应用商店打不开怎么办
    java线程池系列(1)-ThreadPoolExecutor实现原理
    java并发之SynchronousQueue实现原理
    java并发等待条件的实现原理(Condition)
    轻松掌握java读写锁(ReentrantReadWriteLock)的实现原理
    Java NIO 内存映射文件
    Java NIO Path接口和Files类配合操作文件
    Java 序列化 序列化与单例模式 [ 转载 ]
    Java 序列化 JDK序列化总结
    Java 泛型 Java使用泛型的意义
  • 原文地址:https://www.cnblogs.com/yjbjingcha/p/7157772.html
Copyright © 2011-2022 走看看