Message Flood

本文探讨了手机应用在日常生活中的便利性,特别是新年期间发送祝福消息的问题。通过一个实际场景,展示了如何通过编程解决发送祝福消息时遇到的困扰,包括如何高效地管理好友名单和消息接收者列表,避免重复发送消息,从而减轻用户的负担。

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Description

Well, how do you feel about mobile phone? Your answer would probably be something like that "It's so convenient and benefits people a lot". However, If you ask Merlin this question on the New Year's Eve, he will definitely answer "What a trouble! I have to keep my fingers moving on the phone the whole night, because I have so many greeting message to send!" Yes, Merlin has such a long name list of his friends, and he would like to send a greeting message to each of them. What's worse, Merlin has another long name list of senders that have sent message to him, and he doesn't want to send another message to bother them Merlin is so polite that he always replies each message he receives immediately). So, before he begins to send message, he needs to figure to how many friends are left to be sent. Please write a program to help him. Here is something that you should note. First, Merlin's friend list is not ordered, and each name is alphabetic strings and case insensitive. These names are guaranteed to be not duplicated. Second, some senders may send more than one message to Merlin, therefore the sender list may be duplicated. Third, Merlin is known by so many people, that's why some message senders are even not included in his friend list.

Input

There are multiple test cases. In each case, at the first line there are two numbers n and m (1<=n,m<=20000), which is the number of friends and the number of messages he has received. And then there are n lines of alphabetic strings(the length of each will be less than 10), indicating the names of Merlin's friends, one per line. After that there are m lines of alphabetic strings, which are the names of message senders. The input is terminated by n=0.

Output

For each case, print one integer in one line which indicates the number of left friends he must send.

Sample Input

5 3
Inkfish
Henry
Carp
Max
Jericho
Carp
Max
Carp
0

Sample Output

3

 

 

 

题意:先是输入n个好友,然后再是m个给他发短信的人,其中有可能不是他的好友,然后找有多少个好友没有给他发信息

过程:用的map,因为大小一样,所以先将大小写全部转化为大写,然后判断m个里那个存在,只要》=1既存在,如果查到一个则将其附为0,防止下次查找到重复的名字,然后将总人数减1,即可求出。

代码实现:

#include <stdio.h>
#include <string.h>
#include <map>
#include<string>
#include <iostream>
#include <algorithm>
using namespace std;

map<string,int>::iterator it;
char str[101];
int main()
{
    int n,m,i,j,k;
    while(~scanf("%d",&n)&&n)
    {
        map<string,int>q;
        scanf("%d",&m);
        for(i=0; i<n; i++)
        {
            scanf("%s",str);
            k=strlen(str);
            for(j=0;j<k;j++)
            {
                if(str[j]>='a'&&str[j]<='z')
                {
                    str[j]-=32;
                }
            }
            q[str]++;
        }
        for(i=0; i<m; i++)
        {
            scanf("%s",str);
            k=strlen(str);
            for(j=0;j<k;j++)
            {
                if(str[j]>='a'&&str[j]<='z')
                {
                    str[j]-=32;
                }
            }
            if(q[str]!=0)
            {
                n--;
                q[str]=0;
            }
        }
        printf("%d\n",n);
    }
    return 0;
}


 

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