Codeup 6168 问题 A: Speech Patterns (25)

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题目链接：http://codeup.cn/problem.php?cid=100000599&pid=0

题目描述
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?

输入
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].

输出
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.

样例输入
Can1: “Can a can can a can? It can!”

样例输出
can 5

代码

#include <iostream>
#include <string> 
#include <map>using namespace std;bool check(char c) {				//检查字符是否合法if(c >= '0' && c <= '9')return true;if(c >= 'a' && c <= 'z')return true;if(c >= 'A' && c <= 'Z')return true;return false;
}int main() {map<string, int> count;string str;getline(cin, str);unsigned int i = 0;while(i < str.length()) {string word;while(i < str.length() && check(str[i]) == true) { 	//将单词提取到word中if(str[i] >= 'A' && str[i] <= 'Z')				//不区分大小写，把大写统一转化成小写str[i] += 32;word += str[i];i++;}if(word != "") {						 //若单词不为空if(count.find(word) == count.end())  //统计此单词出现的个数count[word] = 1;elsecount[word]++;}while(i < str.length() && check(str[i]) == false)//跳过非法的字符i++;}string ans;		//存放出现最多的单词int max = 0;	//存放最多单词的次数for(map<string, int>::iterator it = count.begin(); it != count.end(); it++) //依次寻找if(it->second > max) {max = it->second;ans = it->first;}cout<<ans<<" "<<max<<endl;return 0;
}