CodeForces - 4C Registration system【STL】【水题】

最新推荐文章于 2019-10-02 20:45:53 发布

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本文介绍了一个名为Berlandesk的邮件服务注册系统的实现方案。该系统通过追加数字来确保用户名唯一，并使用哈希表来跟踪已注册的用户名。文章提供了完整的代码示例，演示了如何响应用户的注册请求。

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【题目描述】
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 32 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

【样例输入】
6
first
first
second
second
third
third

【样例输出】
OK
first1
OK
second1
OK
third1

题目链接：https://codeforces.com/contest/4/problem/C

怎么以前的C题都这么水呢。。。

代码如下：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    string s;
    unordered_map<string,int> mp;
    for(int i=1;i<=n;i++)
    {
        cin>>s;
        if(mp[s]==0) cout<<"OK"<<endl;
        else cout<<s<<mp[s]<<endl;
        mp[s]++;
    }
    return 0;
}