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
//输出出现次数最多的单词,如果有多个,则输出字典序最小的单词 #include <iostream> #include <cstdio> #include <vector> #include <string> #include <cstring> #include <map> #include <unordered_map> using namespace std; string str; unordered_map<string, int> mp; int main(void) { getline(cin, str); int length = str.size(); for (int i = 0; i < length; ++i) { if (str[i] >= 'A' && str[i] <= 'Z')//首先大写字母换成小写字母 str[i] += 32; } int i = 0; string word; while (i < length) { word.clear(); while (i < length) { if ((str[i] <= 'z' && str[i] >= 'a') || (str[i] <= '9' && str[i] >= '0')) break; ++i; } while (i < length) { if ((str[i] <= 'z' && str[i] >= 'a') || (str[i] <= '9' && str[i] >= '0')) word.push_back(str[i]); else break; ++i; }//其实这里有一个问题,就是最后的时候可以存在空word存入到mp中 if (!word.empty()) ++mp[word]; } unordered_map<string, int>::iterator wordIter = mp.begin(); for (unordered_map<string, int>::iterator iter = mp.begin(); iter != mp.end(); ++iter) { if (iter -> second > wordIter -> second || (iter -> second == wordIter -> second && iter -> first < wordIter -> first)) wordIter = iter; } cout << wordIter -> first << " " << wordIter -> second; return 0; }//输出出现次数最多的单词,如果有多个,则输出字典序最小的单词 #include <iostream> #include <cstdio> #include <vector> #include <string> #include <cstring> #include <map> #include <unordered_map> using namespace std; string str; unordered_map<string, int> mp; int main(void) { getline(cin, str); int length = str.size(); for (int i = 0; i < length; ++i) str[i] = tolower(str[i]); int i = 0; string word; while (i < length) { word.clear(); while (i < length && !isalnum(str[i])) ++i; while (i < length && isalnum(str[i])) { word.push_back(str[i]); ++i; }//其实这里有一个问题,就是最后的时候可以存在空word存入到mp中 if (!word.empty()) ++mp[word]; } unordered_map<string, int>::iterator wordIter = mp.begin(); for (unordered_map<string, int>::iterator iter = mp.begin(); iter != mp.end(); ++iter) { if (iter -> second > wordIter -> second || (iter -> second == wordIter -> second && iter -> first < wordIter -> first)) wordIter = iter; } printf("%s %d\n", wordIter -> first.c_str(), wordIter -> second); return 0; }

本文介绍了一种通过分析个人文本样本中词汇偏好来确定其身份的方法。这种方法对于在线环境中验证同一用户的身份非常有用,比如判断不同的网络账号是否属于同一人。通过对输入文本进行预处理并统计各单词出现频率,可以找出最常使用的单词及其出现次数。
2014

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