• Codeforce:4C. Registration system (映射)


    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 name i 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
    

    思路:这道题是一道映射题,name对应的有几个即可

    #include<bits/stdc++.h>
    using namespace std;
    int main() {
    	//freopen("in.txt","r",stdin);
    	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    	int t; cin >> t;
    	unordered_map<string, int>s; string str;//unordered_map 不进行排序优化时间,不过仅从900+ms减少到600+ms
    	while (t--) {
    		cin >> str; 
    		s[str]++;
    		if (s[str] > 1)cout << str << s[str] - 1 << endl;
    		else cout << "OK" << endl;
    	}
    }
    
    //dalao的优化时间算法:154ms
    #include<bits/stdc++.h>
    using namespace std;
    #define getchar_unlocked() _getchar_nolock()
    template<typename T> inline bool sc(T& num) { 
    	bool neg = 0; int c; num = 0; 
    	while (c = getchar_unlocked(), c < 33) { 
    		if (c == EOF) return false; 
    	} 
    	if (c == '-') { 
    		neg = 1; c = getchar_unlocked();
    	} 
    	for (; c > 47; c = getchar_unlocked()) num = num * 10 + c - 48; if (neg) num *= -1; return true; }
    template<typename T, typename ...Args> inline void sc(T& num, Args&...args) { bool neg = 0; int c; num = 0; while (c = getchar_unlocked(), c < 33) { ; } if (c == '-') { neg = 1; c = getchar_unlocked(); } for (; c > 47; c = getchar_unlocked()) num = num * 10 + c - 48; if (neg) num *= -1; sc(args...); }
    inline void getstr(string& str) { 
    	str.clear(); char cur; 
    	while (cur = getchar_unlocked(), cur < 33) { ; }
    	while (cur > 32) { str += cur; cur = getchar_unlocked(); } 
    }
    
    int32_t main()
    {
    	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    	int n;
    	sc(n);
    	unordered_map<string, int> arr;
    	string str;
    	for (int i = 0; i < n; i++)
    	{
    		getstr(str);
    		int r = arr[str]++;
    		if (!r)
    		{
    			cout << "OK\n";
    		}
    		else
    		{
    			cout << str << r << '\n';
    		}
    	}
    
    	return 0;
    }
    
  • 相关阅读:
    EF实体框架处理实体之间关联关系与EF延迟机制(下)
    EF Code First Migrations数据库迁移
    EF的默认映射以及如何使用Data Annotations和Fluent API配置数据库的映射
    学习Entity Framework 中的Code First
    使用Visual Studio Code开发Asp.Net Core WebApi学习笔记(四)-- Middleware
    学习ASP.NET Core,你必须了解无处不在的“依赖注入”
    EF Code First 学习笔记:表映射 多个Entity到一张表和一个Entity到多张表
    OWIN and Katana
    ASP.NET Core 1.0基础之依赖注入
    Mysql支持的数据类型(总结)
  • 原文地址:https://www.cnblogs.com/RioTian/p/13360874.html
Copyright © 2020-2023  润新知