hdu 1242 disney (map模拟)

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本文介绍了一个应用于迪士尼乐园的评分系统，该系统支持用户输入或更新分数、查询平均分及最高分等功能，并通过样例展示了系统的具体操作流程。

disney

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1039 Accepted Submission(s): 299

Problem Description

A new machine was introduced into disney world. Every day, there is a long queue because of its function. Everyone can input their score or update their score. If you want to know the maximum or the average, it can help you, too.What's more, if you don't want that your score be saw, you can delete your register from it. Since the new machine was introduced, everyone be more hard.

Input

The input contain only one case.
There are kinds of format that show different functions:
(1)NEW name score(you can use it to input your score or update your score)
name: the name of current user.The length of the name is no longer than 8.
score: the score of current user.(0<score<=100)
(2)AVERAGE(you can use is to get the average score)
(3)MAX(you can use is to get the maximal score and whose score is the maximum.)
(4)DELETE name: delete the name and its record too.
name: the name of current user.The length of the name is no longer than 8.
(5)QUIT(this command is used to turn off the machine,is the end of the input)

Output

(1)For "NEW name score" , if it's there is a lack of record of the user. print "A new record".else print "update succeed".
(2)For "AVERAGE" , you should print the average score rounded to two digits after the decimal point.if there are none record in the machine,the average is 0.00;
(3)For "MAX" , you should first print the maximal score and the number that whose score is the maximum, then print all of their names Lexicographicly, and each name owns a single line. If there isn't any record in the machine,just output"0 0".
(4)For "DELETE name" , if there is a lack of record of the user. print "no such record".else print "delete succeed".

The population of disney is no more than 100.

Sample Input

      
       NEW mickey 99
NEW mini 88
MAX
AVERAGE
DELETE winnie
QUIT

Sample Output

      
       A new record
A new record
99 1
mickey
93.50
no such record

Author

kiki

Source

HDU 2007-11 Programming Contest

思路：用map来模拟。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;

int main()
{
    string n;
    map<string,int> m;
    map<string,int> ::iterator it;
    double sum=0;
    while(cin>>n)
    {
        if(n=="NEW")
        {
            string name;
            int grade;
            cin>>name;
            cin>>grade;
            it=m.find(name);
            if(it==m.end())
            {
                m[name]=grade;
                printf("A new record\n");
                sum+=grade;
            }
            else
            {
                printf("update succeed\n");
                sum-=m[name];
                sum+=grade;
                m[name]=grade;
            }
        }
        else if(n=="AVERAGE")
        {
            if(m.size()==0)
            {
                printf("0.00\n");
            }
            else
            {
                double ave=sum/(m.size()*1.0);
                printf("%.2lf\n",ave);
            }
        }
        else if(n=="MAX")
        {
            if(m.size()==0)
            {
                printf("0 0\n");
            }
            else
            {
                int maxn=m.begin()->second;
                for(it=m.begin();it!=m.end();it++)
                {
                    if(it->second>maxn) maxn=it->second;
                }
                vector<string> f;
                f.clear();
                for(it=m.begin();it!=m.end();it++)
                {
                    if(it->second==maxn)
                    {
                        f.push_back(it->first);
                    }
                }
                sort(f.begin(),f.end());
                printf("%d %d\n",maxn,f.size());
                for(int i=0;i<f.size();i++)
                {
                    cout<<f[i]<<endl;
                }
            }
        }
        else if(n=="DELETE")
        {
            string nam;
            cin>>nam;
            it=m.find(nam);
            if(it==m.end())
            {
                printf("no such record\n");
            }
            else
            {
                printf("delete succeed\n");
                sum-=it->second;
                m.erase(it);
            }
        }
        else break;
    }
    return 0;
}