Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem D - Zipf's Law

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

2

In practice, the difference between theory and practice is always
greater than the difference between theory and practice in theory.
	- Anonymous

Man will occasionally stumble over the truth, but most of the
time he will pick himself up and continue on.
        - W. S. L. Churchill
EndOfText

Output for Sample Input

between
difference
in
will


#include <stdio.h>
#include <map>
#include <iostream>
#include <sstream>
#include <string.h>
using namespace std;
int main() {
    int n, cases = 0;
    int i, m;
    string line;
    while(scanf("%d", &n) == 1) {
        if(cases)   puts("");
        cases++;
        map<string, int> R;
        while(getline(cin, line)) {
            if(line == "EndOfText") break;
            m = line.length();
            for(i = 0; i < m; i++) {
                if(line[i] >= 'a' && line[i] <= 'z'){}
                else if(line[i] >= 'A' && line[i] <= 'Z')
                    line[i] = line[i]+32;
                else
                    line[i] = ' ';
            }
            stringstream sin(line);
            while(sin >> line)
                R[line]++;
        }
        int found = 0;
        for(map<string, int>::iterator it = R.begin();
            it != R.end(); it++) {
            if(it->second == n) {
                cout << it->first << endl;
                found = 1;
            }
        }
        if(!found)
            puts("There is no such word.");
    }
    return 0;
}
/*
2
a-a
EndOfText
*/