CSU-ACM暑假集训基础组七夕专场 D - Problem D

本文介绍了一个简单的邮箱注册系统的实现过程，该系统能够处理重复用户名的情况并自动生成新的用户名。使用C++语言，通过标准输入输出进行交互，并利用map数据结构来存储用户名及其出现次数。

D - Problem D

Time Limit:5000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Description

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 Input

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

map的简单题，代码如下：

#include <iostream>
#include<cstdio>
#include<map>
#include<string>
#include<cstring>
using namespace std;
int n,maxn;
string maxc;
//char* a;
string a;
map<string, int> name;
int main()
{
    scanf("%d",&n);
        int i;
        map<string ,int>::iterator ite;
        for(i=0,ite=name.begin();i<n;i++,ite++)
        {
            //scanf("%s",a);
            cin>>a;
            name[a]++;
            if(name[a]==1) cout<<"OK"<<endl;
           // else printf("%s%d\n",a,name[a]-1);
           else cout<<a<<name[a]-1<<endl;
        }
    return 0;
}