Codeforces Beta Round #5 A. Chat Server's Outgoing Traffic 水题
A. Chat Server's Outgoing Traffic
题目连接:
http://www.codeforces.com/contest/5/problem/A
Description
Polycarp is working on a new project called "Polychat". Following modern tendencies in IT, he decided, that this project should contain chat as well. To achieve this goal, Polycarp has spent several hours in front of his laptop and implemented a chat server that can process three types of commands:
Include a person to the chat ('Add' command).
Remove a person from the chat ('Remove' command).
Send a message from a person to all people, who are currently in the chat, including the one, who sends the message ('Send' command).
Now Polycarp wants to find out the amount of outgoing traffic that the server will produce while processing a particular set of commands.
Polycarp knows that chat server sends no traffic for 'Add' and 'Remove' commands. When 'Send' command is processed, server sends l bytes to each participant of the chat, where l is the length of the message.
As Polycarp has no time, he is asking for your help in solving this problem.
Input
Input file will contain not more than 100 commands, each in its own line. No line will exceed 100 characters. Formats of the commands will be the following:
+ for 'Add' command.
- for 'Remove' command.
<sender_name>:<message_text> for 'Send' command.
and <sender_name> is a non-empty sequence of Latin letters and digits. <message_text> can contain letters, digits and spaces, but can't start or end with a space. <message_text> can be an empty line.
It is guaranteed, that input data are correct, i.e. there will be no 'Add' command if person with such a name is already in the chat, there will be no 'Remove' command if there is no person with such a name in the chat etc.
All names are case-sensitive.
Output
Print a single number — answer to the problem.
Sample Input
+Mike
Mike:hello
+Kate
+Dmitry
-Dmitry
Kate:hi
-Kate
Sample Output
9
Hint
题意
有一个聊天室
有三种操作:
1.+ xxx 进来一个人
2.- xxx 出去一个人
3.xxx:xxxxx xxx对着所有人说xxxxx,然后这个会对所有人发送一个长度为l的信息。
问你一共发了多长的信息。
题解:
其实根本不用管究竟是谁进来,谁出去。
反正我只关注人数嘛。
代码
#include<bits/stdc++.h>
using namespace std;
int a,b;
string s;
int main()
{
while(getline(cin,s))
{
if(s[0]=='+')a++;
else if(s[0]=='-')a--;
else b+=a*(s.size()-s.find(':')-1);
}
cout<<b<<endl;
}
Codeforces Beta Round #5 A. Chat Server's Outgoing Traffic 水题的更多相关文章
- Codeforces Beta Round #9 (Div. 2 Only) B. Running Student 水题
B. Running Student 题目连接: http://www.codeforces.com/contest/9/problem/B Description And again a misfo ...
- Codeforces Beta Round #9 (Div. 2 Only) A. Die Roll 水题
A. Die Roll 题目连接: http://www.codeforces.com/contest/9/problem/A Description Yakko, Wakko and Dot, wo ...
- Codeforces Beta Round #3 A. Shortest path of the king 水题
A. Shortest path of the king 题目连接: http://www.codeforces.com/contest/3/problem/A Description The kin ...
- CodeForces.5A Chat Server's Outgoing Traffic
Chat Server's Outgoing Traffic 点我挑战提目 考察点 模拟 字符串 Time Mem Len Lang 30 0 543 c++ 题意分析 给出类似一个群的即时通讯系统, ...
- Chat Server's Outgoing Traffic
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=93359#problem/A (456321) http://codeforces.com ...
- Codeforces Beta Round #80 (Div. 2 Only)【ABCD】
Codeforces Beta Round #80 (Div. 2 Only) A Blackjack1 题意 一共52张扑克,A代表1或者11,2-10表示自己的数字,其他都表示10 现在你已经有一 ...
- Codeforces Beta Round #62 题解【ABCD】
Codeforces Beta Round #62 A Irrational problem 题意 f(x) = x mod p1 mod p2 mod p3 mod p4 问你[a,b]中有多少个数 ...
- Codeforces Beta Round #83 (Div. 1 Only)题解【ABCD】
Codeforces Beta Round #83 (Div. 1 Only) A. Dorm Water Supply 题意 给你一个n点m边的图,保证每个点的入度和出度最多为1 如果这个点入度为0 ...
- Codeforces Beta Round #13 C. Sequence (DP)
题目大意 给一个数列,长度不超过 5000,每次可以将其中的一个数加 1 或者减 1,问,最少需要多少次操作,才能使得这个数列单调不降 数列中每个数为 -109-109 中的一个数 做法分析 先这样考 ...
随机推荐
- isolation forest进行异常点检测
一.简介 孤立森林(Isolation Forest)是另外一种高效的异常检测算法,它和随机森林类似,但每次选择划分属性和划分点(值)时都是随机的,而不是根据信息增益或者基尼指数来选择.在建树过程中, ...
- Machine Learning系列--归一化方法总结
一.数据的标准化(normalization)和归一化 数据的标准化(normalization)是将数据按比例缩放,使之落入一个小的特定区间.在某些比较和评价的指标处理中经常会用到,去除数据的单位限 ...
- Java Eclipse 配置
1.清除多余记录 最近用eclipse打包jar的时候,需要指定一个main函数.需要先运行一下main函数,eclipse的Runnable JAR File Specification 下的Lau ...
- 《深入理解Java虚拟机》笔记--第十二章、Java内存模型与线程
主要内容:虚拟机如何实现多线程.多线程之间由于共享和竞争数据而导致的一系列问题及解决方案. Java内存模型: Java内存模型的主要目标是定义程序中各个变量的访问规则,即在虚拟机中将变量存储 ...
- kvm安装准备
到实际情况下,做虚拟化是直接做在真机上. 但实验时,可以在虚拟机上进行.(因为做实验的时候没办法连接到桥接模式的网络,所以使用了NAT方式来连接网络) 在vmware安装centos 64bit fo ...
- docker swarm join 报错
[peter@minion ~]$ docker swarm join --token SWMTKN-1-3mj5po3c7o04le7quhkdhz6pm9b8ziv3qe0u7hx0hrgxsna ...
- java基础11 继承(super、extends关键字和重写,这三个要素出现的前提:必须存在继承关系)
面向对象的三大特征: 1.封装 (将一类属性封装起来,并提供set()和get()方法给其他对象设置和获取值.或者是将一个运算方法封装起来,其他对象需要此种做运算时,给此对象调用) 2.继承 ...
- mknod命令
mknod - make block or character special filesmknod [OPTION]... NAME TYPE [MAJOR MINOR] option 有用的 ...
- fprintf输出到文件中,sprintf输出到字符串中. 如: fprintf(fp,"%s",name); fp为文件指针 sprintf(buff,"%s",name); buff为字符数组
fprintf输出到文件中,sprintf输出到字符串中. 如: fprintf(fp,"%s",name); fp为文件指针 sprintf(buff,"%s" ...
- golang类型转换小总结
1. int <--> string 1.1. int --> string str := strconv.Itoa(intVal) 当然,整数转换成字符串还有其他方法,比如 fmt ...