PAT

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

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给定条件：
1.一行字符串

要求：
1.字符串中出现最多的子字符串

求解：
1.子字符串是由数字和字母组成 即 子字符串之间是由非数字和字母分开的
2.遍历字符串通过判断当前字符是否是字母或数字得到子字符串
3.利用map记录子字符串出现的次数

需要注意的是：
1.只有一个子字符串
2.最后一个字符是数字或字母
第1种情况是第2种情况的特殊情况

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#include <cstdio>#include <iostream>#include <map>#include <cctype>using namespace std;string s;string t = "";map<string, int> m;string ans;int maxcnt;int main() {getline(cin, s);for(int i = 0; i < s.length(); i++) {if(isalnum(s[i])) {s[i] = tolower(s[i]);t += s[i];} if(!isalnum(s[i]) || i == s.length()-1) {if(t != "") {m[t]++;if(m[t] > maxcnt) {maxcnt = m[t];ans = t;}}t="";}}cout<<ans<<" "<<maxcnt<<endl;return 0;}