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
题意：输入一个字符串，求出现次数最多的单词（字母或数字都可以）的数量以及它出现的次数，如果有并列的话，就输出字典序里面的第一个
思路：
- 输入带有空格的字符串用cin.getline(cin,s)
- isalnum()函数的应用：用来检测一个字符是否是字母或者十进制数字
isalpha() 函数：仅仅检测一个字符是否是字母
isdigit() 函数：仅仅检测一个字符是否是十进制数字
如果一个字符被 isalpha() 或者 isdigit() 检测后返回“真”，那么它被 isalnum() 检测后也一定会返回“真”。 - 将字符串进行遍历，用map<string,int>来存储字符串和字符串出现的次数，当是数字或者字母的时候，将遍历到的这个字符用tolower()转化为小写字母，然后t = t + s[i],之后再判断如果不是字母或者数字的话，如果t不为空，就map[t]++，然后再把字符串t清空，最后遍历一下map把结果求出来即可。
#include<iostream>
#include<cstdio>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<cctype>using namespace std;int main() {
string s,t;getline(cin,s);int length = s.length();map<string,int> map;for(int i = 0;i < length;i++) {
if(isalnum(s[i])) {
s[i] = tolower(s[i]);t = t + s[i];}if(!isalnum(s[i]) || i == length - 1) {
if(t.length() != 0) {
map[t]++; }t = "";}}int maxn = 0;for(auto it = map.begin();it != map.end();it++) {
if(it->second > maxn) {
t = it->first;maxn = it->second;}}cout << t << " " << maxn << endl;return 0;
}