1071. Speech Patterns (25)

最新推荐文章于 2025-01-02 01:02:14 发布

zju_lemon

最新推荐文章于 2025-01-02 01:02:14 发布

阅读量511

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT

本文链接：https://blog.youkuaiyun.com/matrixa17/article/details/19361125

PAT 专栏收录该内容

79 篇文章

订阅专栏

本文深入探讨了深度学习技术在自然语言处理领域的应用，包括词法分析、句法分析、语义分析等关键任务，展示了深度学习如何通过复杂的神经网络结构提升语言理解与生成能力。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when validating, for example, whether it's still the same person behind an online avatar.

Now given a paragraph of text sampled from someone's speech, can you find the person's most commonly used word?

Input Specification:

Each input file contains one test case. For each case, there is one line of text no more than 1048576 characters in length, terminated by a carriage return '\n'. The input contains at least one alphanumerical character, i.e., one character from the set [0-9 A-Z a-z].

Output Specification:

For each test case, print in one line the most commonly occurring word in the input text, followed by a space and the number of times it has occurred in the input. If there are more than one such words, print the lexicographically smallest one. The word should be printed in all lower case. Here a "word" is defined as a continuous sequence of alphanumerical characters separated by non-alphanumerical characters or the line beginning/end.

Note that words are case insensitive.

Sample Input:

Can1: "Can a can can a can?  It can!"

Sample Output:

can 5

思路：用map<string,int>来储存并计数找到最大出现次数最大的那个就是OK 没有什么坑点！

AC参考代码：

#include <iostream>
#include <map>
using namespace std;
string lowerIt(string s){
    for(int i=0;i<s.size();i++){
        if(s[i]>='A'&&s[i]<='Z'){
            s[i] = s[i]-'A'+'a';
        }
    }
    return s;
}
map<string,int> myMap;
int main()
{
    string s;
    getline(cin,s);
    string temp = "";
    for(int i=0;i<s.size();i++){
        if((s[i]>='0'&&s[i]<='9') || (s[i]>='a'&&s[i]<='z') || (s[i]>='A'&&s[i]<='Z')){
            temp +=s[i];
        }else if(temp!=""){
            temp = lowerIt(temp);
            myMap[temp]++;
            temp = "";
        }
        if((i==s.size()-1) && ((s[i]>='0'&&s[i]<='9') || (s[i]>='a'&&s[i]<='z') || (s[i]>='A'&&s[i]<='Z'))){
           if(temp!=""){
                temp = lowerIt(temp);
                myMap[temp]++;
                temp = "";
           }
        }
    }
    map<string,int>::iterator it;
    string result = (*myMap.begin()).first;
    int num = (*myMap.begin()).second;
    for(it=myMap.begin();it!=myMap.end();it++){
        if((*it).second>num){
            num = (*it).second;
            result = (*it).first;
        }
    }
    cout<<result<<" "<<num;
    return 0;
}