Zipf's Law（map容器）

Zipf's Law

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 1767 Accepted: 532

Description

Harvard linguistics professor George Kingsley Zipf (1902-1950) observed that the frequency of the kth most common word in a text is roughly proportional to 1/k. He justified his observations in a book titled Human behavior and the principle of least effort published in 1949. While Zipf's rationale has largely been discredited, the principle still holds, and others have afforded it a more sound mathematical basis.
You are to find all the words occurring n times in an English text. A word is a sequence of letters. Words are separated by non-letters. Capitalization should be ignored. A word can be of any length that an English word can be.

Input

Input consists of several test cases. The first line of each case contains a single positive integer n. Several lines of text follow which will contain no more than 10000 words. The text for each case is terminated by a single line containing EndOfText. EndOfText does not appear elsewhere in the input and is not considered a word.

Output

For each test case, output the words which occur n times in the input text, one word per line, lower case, in alphabetical order. If there are no such words in input, output the following line:
There is no such word.

Leave a blank line between cases.

Sample Input

2In practice, the difference between theory and practice is alwaysgreater than the difference between theory and practice in theory.- AnonymousMan will occasionally stumble over the truth, but most of thetime he will pick himself up and continue on.        - W. S. L. ChurchillEndOfText

Sample Output

betweendifferenceinwill

Source

Waterloo local 2001.06.02

map容器，很轻松啊！

 

#include<cstdio>#include<iostream>#include<string>#include<sstream>#include<cstring>#include<map>using namespace std;int main(){    string s,buf;    int i,f,n,num=0;    while(cin>>n)    {        f=0;        map<string,int>mp;        while (cin>>s&&s!="EndOfText")        {             for(i=0;i<s.length();i++)             {                  if (isalpha(s[i]))                  {                      s[i]=tolower(s[i]);                  }                  else                  {                      s[i]=' ';                  }             }              stringstream ss(s);              while (ss>>buf)              {                  mp[buf]++;              }        }        if(num!=0)        {            cout<<endl;        }        for(map<string,int>::iterator it=mp.begin();it!=mp.end();it++)        {            if(it->second==n)            {                cout<<it->first<<endl;                f=1;            }        }        if(f==0)        {            cout<<"There is no such word."<<endl;        }        num++;    }}