【C++】PAT(advanced level)1071. Speech Patterns (25)

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本文介绍了一种用于分析文本中单词使用频率的方法，并通过一个具体的示例演示了如何找出一段文本中最常出现的单词。该方法适用于识别说话者的语言习惯，有助于确认在线身份的真实性和一致性。

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1071. Speech Patterns (25)

时间限制

300 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

HOU, Qiming

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

1.尝试用while(scanf()) 不过注意，一行的结尾是\n2.

2.第一次，如果只有一个字母结尾的那边忘了设置max的值，导致最后一个测试点没过。

3.map用来统计还是不错的，如果时间不是特别苛刻的话。

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


int main(){
	//freopen("in.txt","r",stdin);
	map<string,int> sv;
	map<string,int>::iterator it;
	//input
	char cc;
	string a;
	int max=-1;
	while(scanf("%c",&cc)){
		if(cc=='\n'){
			char ch;
			if(a!=""){
				sv[a]++;
				if(sv[a]>max){
					max=sv[a];
				}
			}
			break;
		}else if((cc<='9'&&cc>='0')||(cc>='A'&&cc<='Z')||(cc>='a'&&cc<='z')){
			if(cc>='A'&&cc<='Z'){
				cc=cc-'A'+'a';
			}
			a.push_back(cc);
		}else{
			if(a!=""){
				sv[a]++;
				if(sv[a]>max){
					max=sv[a];
				}
			}
			a.clear();
		}
	}
	for(it=sv.begin();it!=sv.end();it++){
		if((*it).second==max){
			cout<<(*it).first<<" "<<max;
			break;
		}
	}
	//fclose(stdin);
	system("pause");
	return 0;
}