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;
}

Codeforce:4C. Registration system (映射)的更多相关文章

  1. (水题)Codeforces - 4C - Registration system

    https://codeforces.com/problemset/problem/4/C 用来哈希的一道题目,用map也可以强行过,但是性能慢了6倍,说明是在字符串比较的时候花费了接近6倍的时间. ...

  2. (用了map) Registration system

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=93241#problem/C (654123) http://codeforces.com ...

  3. 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 ...

  4. 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 ...

  5. c题 Registration system

    Description A new e-mail service "Berlandesk" is going to be opened in Berland in the near ...

  6. 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> # ...

  7. nyoj 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 ...

  8. codeforces Registration system

     Registration system A new e-mail service "Berlandesk" is going to be opened in Berland in ...

  9. 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 ...

  10. Registration system

    Registration system 时间限制:1000 ms  |  内存限制:65535 KB 难度:2 描写叙述 A new e-mail service "Berlandesk&q ...

随机推荐

  1. Eclipse 安装 ABAP 插件报错 Microsoft Visual C++ 2013 (x64) 快速解决

    去官网下载Microsoft Visual C++ 2013 (x64) 安装   Download Visual C++ Redistributable Packages for Visual St ...

  2. 有什么BI工具可以实现中国式报表?

    BI(Business Intelligence)工具是指用于帮助企业收集.分析.处理和展示数据的软件工具,以支持企业决策制定和业务运营优化的技术系统. 中国式报表在BI工具中的实现主要涉及到对中国商 ...

  3. 分布式文件系统HDFS简介

    HDFS实现目标: 兼容廉价的硬件设备    支持大数据集   实现流数据读写   支持简单的文件模型    强大的跨平台兼容性 自身的局限性: 不适合低延迟的数据访问   无法高效储存大量小文件  ...

  4. Hudi 在 vivo 湖仓一体的落地实践

    作者:vivo 互联网大数据团队 - Xu Yu 在增效降本的大背景下,vivo大数据基础团队引入Hudi组件为公司业务部门湖仓加速的场景进行赋能.主要应用在流批同源.实时链路优化及宽表拼接等业务场景 ...

  5. OpenGL纹理转换谜团:纹理写入FRAMEBUFFER后的镜像现象

    在OpenGL中,最近将一个 GL_TEXTURE_2D 纹理写入到 GL_FRAMEBUFFER ,然后从GL_FRAMEBUFFER读取为GL_TEXTURE_2D纹理后,发现GL_TEXTURE ...

  6. Windows 无法加载这个硬件的设备驱动程序。驱动程序可能已损坏或不见了。 (代码 39)

    哔站中有视频解决方案,可以直观看如何操作:Windows 无法加载这个硬件的设备驱动程序.驱动程序可能已损坏或不见了. (代码 39) 第一步:明确感叹号故障硬件(我的是蓝牙也可以是别的)--右键&q ...

  7. .NET周刊 【12月第3期 2023-12-24】

    国内文章 CAP 8.0 版本发布通告 - CAP 7岁生日快乐! https://www.cnblogs.com/savorboard/p/cap-8-0.html 今天宣布CAP 8.0版本正式发 ...

  8. 开心自走棋:使用 Laf 云开发支撑数百万玩家

    先介绍一下开心自走棋 开心自走棋是一款剑与魔法的烧脑自走棋游戏.以著名的魔幻世界观为蓝本,采用了轻松可爱的画面风格,精致细腻的动画和特效来还原魔兽之战. 现在市面上自走棋游戏多是 PvP 玩法为主,而 ...

  9. Java 注解的实现原理

    注解的本质 在 java.lang.annotation.Annotation 接口中有这样的描述: The common interface extended by all annotation i ...

  10. 解决QObject::moveToThread: Current thread (0x56059f9b0f70) is not the object's t

    对 opencv 降级 pip install opencv-python==4.1.2.30