PAT Advanced 1071 Speech Patterns (25分)

标签：最大的   turn   ack   cal   down   sed   ase   div   get   

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

这题用头索引和尾索引进行扫描
如果whie尾索引不是字母和数字，尾指针++
然后进行截取字符串（同时进行判断是否是大于最大的，resKey和resValue记录最大的key和最大的value）
while不是字母和数字，进行尾索引++，为了过滤连续的特殊字符
然后使start = _end
最后打印即可

#include <iostream>
#include <string>
#include <unordered_map>
using namespace std;
int main(){
    string str, resKey;
    int resValue;
    unordered_map<string, int> m;
    getline(cin, str);
    for(int i = 0; i < str.length(); i++)
        str[i] = tolower(str[i]);//转换小写字母
    int start = 0;
    for(int _end = 0; _end < str.length(); _end++) {
        string tmp = "";
        while(isalnum(str[_end])) _end++;
        if(++m[str.substr(start, _end - start)] > resValue){
            resKey = str.substr(start, _end - start);
            resValue = m[resKey];
        }
        while(!isalnum(str[_end])) _end++;
        start = _end;
    }
    cout << resKey << " " <<resValue;
    return 0;
}

PAT Advanced 1071 Speech Patterns (25分)

标签：最大的   turn   ack   cal   down   sed   ase   div   get   

原文地址：https://www.cnblogs.com/littlepage/p/12232875.html