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

题意:

AC代码:

 #include<iostream>
#include<cstdio>
#include<string>
#include<cstring> using namespace std;
string str[];
int number[]={}; int em(int k)
{
int i;
for(i=k-;i>;i--)
if(str[k]==str[i]){
number[k]=number[i]+;
cout<<str[k]<<number[k]<<endl;
return -;
}
return ;
} int main()
{
int n;
cin>>n;
int i;
for(i=;i<=n;i++)cin>>str[i];
for(i=;i<=n;i++){if(em(i)==)cout<<"OK"<<endl;}
return ;
}

c题 Registration system的更多相关文章

  1. nyoj 991 Registration system (map)

    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&q ...

  2. ACM 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&q ...

  3. 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 ...

  4. CodeForces-4C Registration system

    // Registration system.cpp : 此文件包含 "main" 函数.程序执行将在此处开始并结束. // #include <iostream> # ...

  5. nyoj 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&q ...

  6. (用了map) Registration system

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=93241#problem/C (654123) http://codeforces.com ...

  7. codeforces Registration system

     Registration system A new e-mail service "Berlandesk" is going to be opened in Berland in ...

  8. Codeforces Beta Round #4 (Div. 2 Only) C. Registration system【裸hash/map】

    C. Registration system time limit per test 5 seconds memory limit per test 64 megabytes input standa ...

  9. Registration system

    Registration system 时间限制:1000 ms  |  内存限制:65535 KB 难度:2 描写叙述 A new e-mail service "Berlandesk&q ...

随机推荐

  1. Java 彩色图转灰度图

    1. 方法1 BufferedImage grayImage = new BufferedImage(width, height, colorImage.TYPE_BYTE_GRAY); Graphi ...

  2. c#弱事件(weak event)

    传统事件publisher和listener是直接相连的,这样会对垃圾回收带来一些问题,例如listener已经不引用任何对象但它仍然被publisher引用 垃圾回收器就不能回收listener所占 ...

  3. ElasticSearch(7)-排序

    引用自ElaticSearch权威指南 一.排序 相关性排序 默认情况下,结果集会按照相关性进行排序 -- 相关性越高,排名越靠前. 这一章我们会讲述相关性是什么以及它是如何计算的. 在此之前,我们先 ...

  4. Chapter 16_5 单一方法

    当一个对象只有一个方法时,可以不用创建接口table,但是要将这个单独的方法作为对象来返回.可以参考迭代器那一节,是如何构造一个迭代器函数,那个函数将状态保存为closure. 一个具有状态的迭代器是 ...

  5. eclipse 导入tomcat7源码

    导入tomcat的源码其实说简单也不简单,说不简单也简单,主要还是环境问题,中间花费了我很多时间,网上找了很多都没什么用,参考一些文章,然后自己慢慢摸索出来的. 环境:(1)jdk:jdk1.6.0_ ...

  6. MocorDroid编译工程快速建立编译环境

    function sprdLunch(){    declare -a arrProj    arrProj=`find out/target/product -name previous_build ...

  7. ps -ef |grep 输出的具体含义

    [root@localhost ~]# ps -ef | grep ApacheJetspeed root     18887 18828  0 08:09 pts/0    00:00:00 gre ...

  8. Sql Server中三种字符串合并方法的性能比较

    文章来自:博客园-DotNet菜园 最近正在处理一个合并字符吕的存储过程,在一个测试系统的开发中,要使用到字符串合并功能,直接在Sql中做.示例:有表內容﹕名称  內容1     abc1      ...

  9. c# 索引器方法

    索引器方法允许我们构建能够以类似访问数组的语法来访问内部子类型的自定义类型 在语法上索引器方法和属性的定义很类似,一样是使用get,set,不同的是索引器是使用this[]创建的. 一个简单的索引器代 ...

  10. linux 目录及文件的命名规则、ls操作

    linux 命名: 1 不超过255个字符 2 严格区分大小写 3 除/外,其他的字符都是合法的 注意:1)避免文件名首字符使用+ - .(避免和隐藏文件混淆) 2)避免文件名使用空格,制表符以及@# ...