NYOJ 题目991 Registration system（set）

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本文详细阐述了在Berland即将开放的电子邮件服务Berlandesk中，如何帮助网站管理员快速启动项目并实现注册系统的原型设计。系统通过接收用户姓名请求，判断其是否存在数据库中，若不存在则插入并回复成功注册的确认信息；反之，则生成新的唯一用户名并插入数据库，同时告知用户新用户名。

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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.

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

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上传者

TCM_张鹏

好吧，看的别人的

ac代码

#include<stdio.h>
#include<set>
#include<string.h>
#include<string>
#include<iostream>
using namespace std;
char *fun(int num)
{
	char s[100010];
	sprintf(s,"%d",num);
	return s;
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		string s,tem;
		int i,j;
		set<string>m;
		for(i=0;i<n;i++)
		{
			cin>>s;
			if(m.count(s)==0)
			{
				m.insert(s);
				printf("OK\n");
			}
			else
			{
				for(j=1;;j++)
				{
					tem=s;
					if(m.count(tem.append(fun(j)))==0)
					{
						//printf("%s\n",s);
						cout<<tem<<endl;
						m.insert(tem);
						break;
					}
				}
			}
		}
	}
}