C. Registration system
time limit per test

5 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

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.

Examples
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

【代码】:

#include<bits/stdc++.h>
using namespace std;
int n,m,cnt;
string s;
map<string,int> mp;
int main()
{
while(cin>>n)
{
while(n--)
{
cin>>s;
mp[s]++;
if(mp[s]==)
puts("OK");
else
cout<<s<<mp[s]-<<endl;
}
}
return ;
}

STL

Codeforces Beta Round #4 (Div. 2 Only) C. Registration system【裸hash/map】的更多相关文章

  1. Codeforces Beta Round #4 (Div. 2 Only) C. Registration system hash

    C. Registration system Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset ...

  2. Codeforces Beta Round #80 (Div. 2 Only)【ABCD】

    Codeforces Beta Round #80 (Div. 2 Only) A Blackjack1 题意 一共52张扑克,A代表1或者11,2-10表示自己的数字,其他都表示10 现在你已经有一 ...

  3. Codeforces Beta Round #83 (Div. 1 Only)题解【ABCD】

    Codeforces Beta Round #83 (Div. 1 Only) A. Dorm Water Supply 题意 给你一个n点m边的图,保证每个点的入度和出度最多为1 如果这个点入度为0 ...

  4. Codeforces Beta Round #79 (Div. 2 Only)

    Codeforces Beta Round #79 (Div. 2 Only) http://codeforces.com/contest/102 A #include<bits/stdc++. ...

  5. Codeforces Beta Round #77 (Div. 2 Only)

    Codeforces Beta Round #77 (Div. 2 Only) http://codeforces.com/contest/96 A #include<bits/stdc++.h ...

  6. Codeforces Beta Round #76 (Div. 2 Only)

    Codeforces Beta Round #76 (Div. 2 Only) http://codeforces.com/contest/94 A #include<bits/stdc++.h ...

  7. Codeforces Beta Round #75 (Div. 2 Only)

    Codeforces Beta Round #75 (Div. 2 Only) http://codeforces.com/contest/92 A #include<iostream> ...

  8. Codeforces Beta Round #74 (Div. 2 Only)

    Codeforces Beta Round #74 (Div. 2 Only) http://codeforces.com/contest/90 A #include<iostream> ...

  9. Codeforces Beta Round #73 (Div. 2 Only)

    Codeforces Beta Round #73 (Div. 2 Only) http://codeforces.com/contest/88 A 模拟 #include<bits/stdc+ ...

随机推荐

  1. Android学习笔记(一)之仿正点闹钟时间齿轮滑动的效果

    看到正点闹钟上的设置时间的滑动效果非常好看,自己就想做一个那样的,在网上就开始搜资料了,看到网上有的齿轮效果的代码非常多,也非常难懂,我就决定自己研究一下,现在我就把我的研究成果分享给大家.我研究的这 ...

  2. python - web自动化测试 - 元素操作 - 鼠标键盘

    # -*- coding:utf-8 -*- ''' @project: web学习 @author: Jimmy @file: 鼠标操作.py @ide: PyCharm Community Edi ...

  3. Python 黑魔法(持续收录)

    Python 黑魔法(持续收录) zip 对矩阵进行转置 a = [[1, 2, 3], [4, 5, 6]] print(list(map(list, zip(*a)))) zip 反转字典 a = ...

  4. static_cast AND dynamic_cast

    类型转换是一种机制,让程序员能够暂时或永久性改变编译器对对象的解释.注意,这并不意味着程序员改变了对象本身,而只是改变了对对象的解释. 在很多情况下,类型转换是合理的需求,可解决重要的兼容问题.因此, ...

  5. UVALive 5029 字典树

    E - Encoded Barcodes Crawling in process...Crawling failedTime Limit:3000MS    Memory Limit:0KB    6 ...

  6. ci重写 配置文件

    server { listen 80; #listen [::]:80; server_name wangyongshun.xyz www.wangyongshun.xyz; index index. ...

  7. BZOJ5157 [Tjoi2014]上升子序列 【树状数组】

    题目链接 BZOJ5157 题解 我们只需计算每个位置为开头产生的贡献大小,就相当于之后每个大于当前位置的位置产生的贡献 + 1之和 离散化后用树状数组维护即可 要注意去重,后面计算的包含之前的,记录 ...

  8. Tomcat学习笔记(二)

    Servlet浅析 javax.servlet.Servlet是一个接口,所有的Servlet必须实现接口里面的方法. 该接口在tomcat/bin中的servlet-api.jar包中. Servl ...

  9. 《R语言实战》读书笔记--学习张丹日志

    从张丹的日志(http://blog.fens.me/rhadoop-r-basic/)中第九条对象看到R对象的几个总结: 1.内在属性 mode length 所有对象都有的属性 2.外部属性 at ...

  10. 《c程序设计语言》-2.6~2.8

    #include <stdio.h> unsigned setbits(unsigned x, int p, int n, unsigned y) { return (x & (( ...