1071. Speech Patterns (25)

最新推荐文章于 2025-01-02 01:02:14 发布

原创最新推荐文章于 2025-01-02 01:02:14 发布 · 154 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

138 篇文章

订阅专栏

本文介绍了一种从文本样本中识别最常用词汇的方法，并通过示例代码展示了如何实现这一功能。该方法对于分析说话人的身份特征非常有用，特别是在验证在线虚拟形象背后是否为同一人时。

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

用map建立键值对

把能组成一个单词的找出来，然后map++；

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<map>
using namespace std;
int pd(char s){
	if((s>='A'&&s<='Z')||(s>='a'&&s<='z')||(s>='0'&&s<='9')) return 1;
	return 0;
}
int main(){
	char s[1048576+10];
	map<string,int> m;
	gets(s);
	int len=strlen(s);
	int num=0;
	string out="";
	for(int i=0;i<len;i++){
		if(s[i]>='A'&&s[i]<='Z') s[i]+=32;
	}
	//cout<<s<<endl;
	for(int i=0;i<len;){
		if(pd(s[i])){
			char fz[10000];
			int cnt=0;
			string f="";
			for(;i<len&&pd(s[i]);i++){
				f+=s[i];
			//	cout<<s[i];
			}
		//	cout<<" ";
			
		//	cout<<f<<endl;
			m[f]++;
			if(m[f]>num){
				num=m[f];
				out=f;
			}
		}
		else i++;
	}
	cout<<out<<" "<<num;
	return 0;
}