《Registration system》

《Registration system》

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.

Input

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 32 characters, which are all lowercase Latin letters.

Output

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.

Sample test(s)

input

4
abacaba
acaba
abacaba
acab

output

OK
OK
abacaba1
OK

input

6
first
first
second
second
third
third

output

OK
first1
OK
second1
OK
third1

题目分析：这是一个注册系统，如果输入的名字还没有被注册，那么返回“OK”；否则返回“输入的名字+已经出现的次数”；

一开始运用的 是循环比较，既把以前注册过的名字保存到一个数组中，然后让刚输入的与其相比较，看相等的次数是多少；能够解决问题，但提交之后 超时了，后来就用的 map，具体看源代码，挺简单的。map的运用确实不错、

源代码：

#include<iostream>
#include<string>
#include<map>
using namespace std;
map<string,int> m;
string a;
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a;
        if(m[a]==0)
        {
            m[a]++;
            cout<<"OK"<<endl;
        }
        else
        {
            cout<<a<<m[a]<<endl;
            m[a]++;

        }
    }
}