【PAT】【Advanced Level】1071. Speech Patterns (25)

最新推荐文章于 2024-05-19 20:54:40 发布

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最新推荐文章于 2024-05-19 20:54:40 发布

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本文分析了PAT A 1071题目的解决方案，该题目要求从给定的文本中找出最常出现的单词，并提供了一种使用C++实现的具体方法。通过对输入文本的逐字符处理，利用map来统计每个单词出现的次数。

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1071. Speech Patterns (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HOU, Qiming

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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1071

https://www.nowcoder.com/pat/5/problem/4120

思路：

字符串处理

逐位读入，每当产生了一个字符串，map进行映射，没有出现过的压入vector、map中设为1，出现过的直接map中+1。

处理完输入之后对vector，依照map中映射的数目进行排序，最后取出最大的

pat上险些超时。。

用char*可能会快一点。。

CODE：

#include<cstring>
#include<iostream>
#include<string>
#include<cstdio>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;

vector<string> vec;

map<string,int> ma;

bool cmp(string a,string b)
{
	if (ma[a]==ma[b])
		return a<b;
	else
		return ma[a]>ma[b];
}

int main()
{
	string t;
	char c;
	t="";
	while (c=getchar())
	{
		//cout<<c;
		if (c>='A'&&c<='Z')
		{
			c+=32;
		}
		if ((c>='a'&&c<='z')||(c>='0'&&c<='9'))
		{
			t+=c;
		}
		else
		{
			if (t!="")
			{
				if (ma[t]!=0)
				{
					ma[t]=ma[t]+1;
				}
				else
				{
					ma[t]=1;
					vec.push_back(t);
				}
				t="";
			}
		}
		if (c=='\n') break;
	}
	sort(vec.begin(),vec.end(),cmp);
	printf("%s %d",vec[0].c_str(),ma[vec[0]]);
	return 0;
}