问题 A: Speech Patterns (25)

时间限制: 1 Sec  内存限制: 32 MB
提交: 449  解决: 146
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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

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解题思路

string类型的处理

对于string类型,它是一个对象,我们尽量使用string对象的方法来操纵它,如push_back(), clear()等等,避免直接对其赋值。

map容器的遍历

迭代器遍历:

for(auto it: mp){} 。

下标访问是不行的。因为下标为string 则没有有序表的规律来对其进行遍历。

AC代码

#include 
#include 
#include 
using namespace std;
int main(){
	map mp;
	string ss, str;
	getline(cin, ss);
	ss+=".";
	//cout << ss<< endl;
	for(int i=0; i::iterator max=mp.begin();
	for(auto it=mp.begin(); it!=mp.end(); it++){
        if(max->second < it->second) max = it;
	}
	cout<< max->first << " " << max->second <

 

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