nyoj Registration system
Registration system
- 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
#include <iostream>
#include <map>
using namespace std;int main()
{
int n;
map<string,int> m;
cin>>n;
while(n--)
{
string s;
cin>>s;if(m[s])
cout<<s<<m[s]<<endl;
else
cout<<"OK"<<endl;
m[s]++;
}
return 0;
}map:数据的插入
在构造map容器后,我们就可以往里面插入数据了。这里讲三种插入数据的方法:
第一种:用insert函数插入pair数据
map<int, string> mapStudent;
mapStudent.insert(pair<int, string>(1,“student_one”));第二种:用insert函数插入value_type数据
map<int, string> mapStudent;
mapStudent.insert(map<int, string>::value_type (1,"student_one"));mapStudent.insert(make_pair(1, "student_one"));
第三种:用数组方式插入数据map<int, string> mapStudent;
mapStudent[1] = “student_one”;
mapStudent[2] = “student_two”;/*如果是
#include <map>
map<string, int> mapStudent;string s;
插入就用m[s]++;*/以上三种用法,虽然都可以实现数据的插入,但是它们是有区别的,当然了第一种和第二种在效果上是完成一样的,用insert函数插入数据,在数据的插入上涉及到集合的唯一性这个概念,即当map中有这个关键字时,insert操作是不能再插入这个数据的,但是用数组方式就不同了,它可以覆盖以前该关键字对应的值,即:如果当前存在该关键字,则覆盖改关键字的值,否则,以改关键字新建一个key—value;
nyoj Registration system的更多相关文章
- nyoj 911 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 ...
- 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 ...
- 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 ...
- 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 ...
- c题 Registration system
Description A new e-mail service "Berlandesk" is going to be opened in Berland in the near ...
- 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> # ...
- (用了map) Registration system
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=93241#problem/C (654123) http://codeforces.com ...
- codeforces Registration system
Registration system A new e-mail service "Berlandesk" is going to be opened in Berland in ...
- 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 ...
随机推荐
- 『科学计算』L0、L1与L2范数_理解
『教程』L0.L1与L2范数 一.L0范数.L1范数.参数稀疏 L0范数是指向量中非0的元素的个数.如果我们用L0范数来规则化一个参数矩阵W的话,就是希望W的大部分元素都是0,换句话说,让参数W是稀 ...
- LeetCode 318. Maximum Product of Word Lengths (状态压缩)
题目大意:给出一些字符串,找出两个不同的字符串之间长度之积的最大值,但要求这两个字符串之间不能拥有相同的字符.(字符只考虑小写字母). 题目分析:字符最多只有26个,因此每个字符串可以用一个二进制数来 ...
- Oracle性能诊断艺术-读书笔记(范围分区)
1. PARTITION RANGE SINGLE 注意:操作2 中的 TABLE ACCESS FULL 并不是全表扫描,只是对分区1 做 全分区扫描 case2 2. 分区范围迭代(PARTITI ...
- 生成输出 URL(16.2)
1.在视图中生成输出 URL 几乎在每一个 MVC 框架应用程序中,你都会希望让用户能够从一个视图导航到另一个视图 —— 通常的做法是在第一个视图中生成一个指向第二个视图的链接,该链接以第二个视图的动 ...
- OAF SubTabLayoutBean隐藏子控件
SubLayout隐藏子控件有两种方法 OASubTabLayoutBean layBean = (OASubTabLayoutBean) webBean.findIndexedChildRecurs ...
- C# Winform 中如何获取本机安装输入法,并设置为默认输出语言,如何打开搜狗输入法和手写板
一.问题: 今天,我整理了一下两个问题 1.如何获取本机安装所有输入法,并设置为系统输出语言 2.如何打开搜狗拼音输入法工具栏和手写板: 二.解决方法 比如:我们要设置搜狗输入法为本机输入语言,要怎么 ...
- Oracle/MySQL decimal/int/number 转字符串
有时客户需要流水数据,当导出为excel的时候,客户编号等很长数字的栏位,被excel变成科学记数法,无法正常查看. 因此,需要将Oracle/MySQL中的decimal/int 转 varchar ...
- python字典{}大括号
#字典 info = { 'name1':'jack', 'name2':'rose', 'name3':'tom', 'name4':'jerry', 'name5':'james' } info[ ...
- 面试题2:单例模式Singleton
首先,单例模式使类在程序生命周期的任何时刻都只有一个实例, 然后,单例的构造函数是私有的,外部程序如果想要访问这个单例类的话, 必须通过 getInstance()来请求(注意是请求)得到这个单例类的 ...
- JQ延迟对象
延迟对象初识 技术一般水平有限,有什么错的地方,望大家指正. ES6已经实现了延迟对象Promise,但是今天主角是JQ里面的延迟对象,套路其实都是差不多的.下面先看一个比较牵强的例子: <bu ...