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;
}
Codeforce:4C. Registration system (映射)的更多相关文章
- (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. ...
- (用了map) Registration system
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=93241#problem/C (654123) http://codeforces.com ...
- 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> # ...
- 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 ...
- 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 ...
- 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
- [ABC266G] Yet Another RGB Sequence
Problem Statement You are given integers $R$, $G$, $B$, and $K$. How many strings $S$ consisting of ...
- [ABC262B] Triangle (Easier)
Problem Statement You are given a simple undirected graph with $N$ vertices and $M$ edges. The verti ...
- keycloak~对接login-status-iframe页面判断用户状态变更
上次我们说了,keycloak的login-status-iframe页面的作用,并解决了跨域情况下,iframe与主页面数据传递的方法,这一次,我们主要分析login-status-iframe.h ...
- springboot去除内嵌tomcat
springboot去除内嵌tomcat步骤 在pom文件中加入以下代码 点击查看代码 <!-- 多模块排除内置tomcat --> <dependency> <grou ...
- 冲刺秋招之牛客刷Java记录第二天
第一题 下列代码输入什么? public class Test { public static Test t1 = new Test(); { System.out.println("blo ...
- SpringBoot-MybatisPlus-Dynamic(多数据源)-springboot-mybatisplus-dynamic-duo-shu-ju-yuan-
title: SpringBoot-MybatisPlus-Dynamic(多数据源) date: 2021-05-07 13:58:06.637 updated: 2021-12-26 17:43: ...
- Pytest07-pytest.ini配置文件
1.pytest配置文件 固定名称:pytest.ini 作用域:当前目录及子目录 具体配置功能见下: [pytest] # 01 把命令行参数自动添加到这里 addopts = -s -v --ht ...
- DVWA Brute Force(暴力破解)全等级
Brute Force(暴力破解) 目录: Brute Force(暴力破解) 1.Low 2.Medium 3.High 方法1--Burp爆破 方法2--Python脚本爆破 4.Impossib ...
- Ynoi
P4688 [Ynoi2016] 掉进兔子洞 序列,静态,求三个区间的可重集的交的大小,离线,\(n,Q\le 10^5\),3s,500MB 缺乏性质 \(\rightarrow\) bitset ...
- Llama2-Chinese项目:5-推理加速
随着大模型参数规模的不断增长,在有限的算力资源下,提升模型的推理速度逐渐变为一个重要的研究方向.常用的推理加速框架包含lmdeploy.FasterTransformer和vLLM等. 一.lmd ...