NYOJ 991 Registration system

Registration system

时间限制: 1000 ms  |  内存限制: 65535 KB
难度: 2
描述

A new e-mail service "Berlandesk" is going to be opened in Berland in the near future.

 The site administration wants to launch their project as soon as possible, that's why they

 ask you to help. You're suggested to implement the prototype of site registration system. 

The system should work on the following principle.

Each time a new user wants to register, he sends to the system a request with his name.

 If such a name does not exist in the system database, it is inserted into the database, and 

the user gets the response OK, confirming the successful registration. If the name already 

exists in the system database, the system makes up a new user name, sends it to the user 

as a prompt and also inserts the prompt into the database. The new name is formed by the

 following rule. Numbers, starting with 1, are appended one after another to name (name1,

 name2, ...), among these numbers the least i is found so that namei does not yet exist in

 the database.

NYOJ 991 Registration system_第1张图片

输入
The first line contains number n (1 ≤ n ≤ 105). The following n lines contain the requests to the system. Each request is a non-empty line, and consists of not more than 1000 characters, which are all lowercase Latin letters.
输出
Print n lines, which are system responses to the requests: OK in case of successful registration, or a prompt with a new name, if the requested name is already taken.
样例输入
4
abacaba
acaba
abacaba
acab
样例输出
OK
OK
abacaba1
OK

来源

/*题解:

STL的映射容器map的使用使解题简单

*/

#include<iostream>
#include<map>
using namespace std;
int main()
{
    int n;
    string str;
    map<string,int>mymap;
    cin>>n;
    while(n--)
	{
        cin>>str;
        if(mymap[str])
        cout<<str<<mymap[str]<<endl;
        else
        cout<<"OK"<<endl;
        
        mymap[str]++;
    }
    return 0;
}
//map是STL的映射容器
//头文件:#include<map>
//对于map<string,int>mymap,string是键值,int是数据 。 
//例如:map["happy"]=6,string str="happy",那么map[str]=6;
test1:

#include<iostream>
#include<map>
using namespace std;
int main()
{
	map<string,int>mymap;//string键值,int数据
	string str = "hellow";
	cout<<mymap[str]<<endl;//初值为零 
}
结果:


test2:

#include<iostream>
#include<map>
using namespace std;
int main()
{
	map<string,int>mymap;//string键值,int数据
	string str = "hellow";
	mymap["hellow"] = 5;
	cout<<mymap[str]<<endl;
}
结果:




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