Codeforces Round #459 (Div. 2):B. Radio Station
B. Radio Station
time limit per test2 seconds
memory limit per test256 megabytes
Problem Dsecription
As the guys fried the radio station facilities, the school principal gave them tasks as a punishment. Dustin’s task was to add comments to nginx configuration for school’s website. The school has n servers. Each server has a name and an ip (names aren’t necessarily unique, but ips are). Dustin knows the ip and name of each server. For simplicity, we’ll assume that an nginx command is of form “command ip;” where command is a string consisting of English lowercase letter only, and ip is the ip of one of school servers.
Each ip is of form “a.b.c.d” where a, b, c and d are non-negative integers less than or equal to 255 (with no leading zeros). The nginx configuration file Dustin has to add comments to has m commands. Nobody ever memorizes the ips of servers, so to understand the configuration better, Dustin has to comment the name of server that the ip belongs to at the end of each line (after each command). More formally, if a line is “command ip;” Dustin has to replace it with “command ip; #name” where name is the name of the server with ip equal to ip.
Dustin doesn’t know anything about nginx, so he panicked again and his friends asked you to do his task for him.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 1000).
The next n lines contain the names and ips of the servers. Each line contains a string name, name of the server and a string ip, ip of the server, separated by space (1 ≤ |name| ≤ 10, name only consists of English lowercase letters). It is guaranteed that all ip are distinct.
The next m lines contain the commands in the configuration file. Each line is of form “command ip;” (1 ≤ |command| ≤ 10, command only consists of English lowercase letters). It is guaranteed that ip belongs to one of the n school servers.
Output
Print m lines, the commands in the configuration file after Dustin did his task.
Examples
input
2 2
main 192.168.0.2
replica 192.168.0.1
block 192.168.0.1;
proxy 192.168.0.2;
output
block 192.168.0.1; #replica
proxy 192.168.0.2; #main
input
3 5
google 8.8.8.8
codeforces 212.193.33.27
server 138.197.64.57
redirect 138.197.64.57;
block 8.8.8.8;
cf 212.193.33.27;
unblock 8.8.8.8;
check 138.197.64.57;
output
redirect 138.197.64.57; #server
block 8.8.8.8; #google
cf 212.193.33.27; #codeforces
unblock 8.8.8.8; #google
check 138.197.64.57; #server
解题心得:
- 题目说了一大堆其实题意很简单,给你n个ip地址,每个地址有一个对应的名字,然后m次询问,每次询问给你一个ip地址,叫你输出ip地址的名字。当然还有一些乱七八糟的输出要求。
- 就是一个hash,将每个地址用一个数表示,然后将地址和这个数存起来,每次询问的时候直接去找这个数,然后将地址输出就行了。主要的是考验一个读题的能力。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5+100;
map <string ,string> maps;
string s1;
int n,m,tot=0;
struct NODE
{
string s;
long long num;
}node[10000];
int main()
{
cin>>n>>m;
for(int i=0;i<n;i++)
{
cin>>s1;
long long x1,x2,x3,x4,ans =0;
scanf("%lld.%lld.%lld.%lld",&x1,&x2,&x3,&x4);//注意输入的时候控制格式
ans += x1*233;
ans = ans*233+x2;
ans = ans*233+x3;
ans = ans*233+x4;
node[tot].num = ans;
node[tot++].s = s1;
}
for(int i=0;i<m;i++)
{
cin>>s1;
long long x1,x2,x3,x4,ans=0;
scanf("%lld.%lld.%lld.%lld;",&x1,&x2,&x3,&x4);
cout<<s1<<" ";
printf("%lld.%lld.%lld.%lld; #",x1,x2,x3,x4);
ans = x1*233;
ans = ans*233+x2;
ans = ans*233+x3;
ans = ans*233+x4;
for(int i=0;i<tot;i++)
if(node[i].num == ans)
{
cout<<node[i].s<<endl;
break;
}
}
return 0;
}
Codeforces Round #459 (Div. 2):B. Radio Station的更多相关文章
- 【Codeforces Round #459 (Div. 2) B】 Radio Station
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 用map模拟一下映射就好了. [代码] #include <bits/stdc++.h> using namespace ...
- Codeforces Round #459 (Div. 2):D. MADMAX(记忆化搜索+博弈论)
D. MADMAX time limit per test1 second memory limit per test256 megabytes Problem Description As we a ...
- Codeforces Round #459 (Div. 2)
A. Eleven time limit per test 1 second memory limit per test 256 megabytes input standard input outp ...
- Codeforces Round #482 (Div. 2) : Kuro and GCD and XOR and SUM (寻找最大异或值)
题目链接:http://codeforces.com/contest/979/problem/D 参考大神博客:https://www.cnblogs.com/kickit/p/9046953.htm ...
- Codeforces Round #482 (Div. 2) :B - Treasure Hunt
题目链接:http://codeforces.com/contest/979/problem/B 解题心得: 这个题题意就是三个人玩游戏,每个人都有一个相同长度的字符串,一共有n轮游戏,每一轮三个人必 ...
- Codeforces Round #532 (Div. 2):F. Ivan and Burgers(贪心+异或基)
F. Ivan and Burgers 题目链接:https://codeforces.com/contest/1100/problem/F 题意: 给出n个数,然后有多个询问,每次回答询问所给出的区 ...
- Codeforces Round #511 (Div. 2):C. Enlarge GCD(数学)
C. Enlarge GCD 题目链接:https://codeforces.com/contest/1047/problem/C 题意: 给出n个数,然后你可以移除一些数.现在要求你移除最少的数,让 ...
- Codeforces Round #514 (Div. 2):D. Nature Reserve(二分+数学)
D. Nature Reserve 题目链接:https://codeforces.com/contest/1059/problem/D 题意: 在二维坐标平面上给出n个数的点,现在要求一个圆,能够容 ...
- Codeforces Round #459 (Div. 2) D. MADMAX DFS+博弈
D. MADMAX time limit per test 1 second memory limit per test 256 megabytes input standard input outp ...
随机推荐
- Ashx登录
<script type="text/javascript"> window.onload = function () { var url = document.get ...
- linq 读取xml
xml 文件如下: <?xml version="1.0" encoding="utf-8" ?><nodes> <node> ...
- Vue系列(2):Vue 安装
前言:关于页面上的知识点,如有侵权,请看 这里 . 关键词:小白.Vue 安装.Vue目录结构.Vue 构建页面流程 ? 初学者安装 vue 用什么好 大家都知道,学 Vue 最好还是去官网学,官网写 ...
- Angular CLI的简单使用(2)
刚才创建了myApp这个项目,看一下这个项目的文件结构. 项目文件概览 Angular CLI项目是做快速试验和开发企业解决方案的基础. 你首先要看的文件是README.md. 它提供了一些如何 ...
- meterpreter > ps
meterpreter > ps Process List============ PID PPID Name Arch Session User Path --- ---- ---- ---- ...
- Excel2Dataset
//获取用户打开的Excel文档路径 private stringkkk() { OpenFileDialog selectFile = new OpenFileDialog(); selectFil ...
- 为OSSIM添加 ossec的linux agent
1,安装环境 [root@node32 test]# yum groupinstall "Development Tools" -y Installed: byacc.x86_64 ...
- linux 命令——36 diff(转)
diff命令是 linux上非常重要的工具,用于比较文件的内容,特别是比较两个版本不同的文件以找到改动的地方.diff在命令行中打印每一个行的改动.最新版本的diff还支持二进制文件.diff程序的输 ...
- python_48_Python3中字符编码与转码
python3默认是Unicode,不用声明# -*- coding:utf-8 -*-,如果声明则是utf-8 unicode='你好' print('utf-8:',unicode.encode( ...
- javascript同步和异步的区别与实现方式
javascript语言是单线程机制.所谓单线程就是按次序执行,执行完一个任务再执行下一个. 对于浏览器来说,也就是无法在渲染页面的同时执行代码. 单线程机制的优点在于实现起来较为简单,运行环境相对简 ...