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社区首页 >专栏 >【南理oj】991 - Registration system(STL - map & string)

【南理oj】991 - Registration system(STL - map & string)

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FishWang
发布2025-08-26 19:30:07
发布2025-08-26 19:30:07
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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

代码语言:javascript
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4
abacaba
acaba
abacaba
acab

样例输出

代码语言:javascript
代码运行次数:0
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OK
OK
abacaba1
OK

来源 爱生活 上传者 TCM_张鹏

每输入一次字符串,在map中记录一次,如果前面出现,则输出出现次数-1即可。

代码如下:

代码语言:javascript
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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <string>
#include <iostream>
using namespace std;
int main()
{
	map<string,int> ant;
	int n;
	string a;
	scanf ("%d",&n);
	while (n--)
	{
		cin >> a;
		if (ant[a] != 0)
			cout << a << ant[a] << endl;
		else
			printf ("OK\n");
		ant[a]++;
	}
	return 0;
}
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