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 H

Count the eWords

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

The Malfunction Checkers Ltd (MCL) is famous for its extraordinary ability to find faults of a malfunctioning device. Many companies depend on it to find out faults of their newly developed products such as TV, VCR, Radio etc.

Few days ago, the administrator of MCL has come to know that someone has been trying hard to damage MCL's reputation by installing an email virus in the main server of MCL. To track this attempt, the administrator has planned to read all incoming emails. But the employees of MCL are very cautious about their privacy and are opposing the plan. As a result, administrator has to change his plan: instead of reading emails, he is planning write a program to count the words in each email and analyze the result.

What the administrator does not know is that the main server's C compiler is already infected. A virus has tactfully damaged some libraries of the compiler. As a result, the strcmp() function of the compiler works in an unexpected way. The problems are:

• It ignores the letters in the 3rd position during comparison.
• The function decides its return value considering only the starting 5 letters.

You are to find out the output of the program written by the administrator.

Input

Input contains sevaral emails, each ends with the word '#' which should not be counted. The input file is terminated by end of file (EOF).

Each email contains a numbers of lines of words. The words may contains any 'readable' character ( except '#' ) and have maximum length of 20. The number of distinct words is limited to 4100, but any word may appear more than once.

Output

This is an longinput string

sort the this line if is has soft line break

#

this is a set of sit

#

timk tidk tisk tia tiy tiz

#

Set #1:

This 1

an 1

break 1

has 1

if 1

is 2

line 2

longi 1

soft 2

strin 1

the 1

this 1

Set #2:

a 1

is 1

of 1

set 1

sit 1

this 1

Set #3:

tiz 3

tisk 3

Mustaq Ahmed

題目描述有錯啊！！

題目描述有錯啊！！

#include <stdio.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <map>
using namespace std;
int main() {
    string s, o;
    int cases = 0;
    while(1) {
        map<string, pair<int, pair<string, string> > > R;
        map<string, pair<int, pair<string, string> > >::iterator it;
        // ignore, count, first, last
        while(1) {
            if(getline(cin, s)) {
                if(s == "#")    break;
                stringstream sin(s);
                while(sin >> s) {
                    if(s.length() >= 5)
                        s = s.substr(0, 5);
                    o = s;
                    if(s.length() >= 3)
                        o[2] = '$';
                    if(R.find(o) == R.end())
                        R[o] = make_pair(1, make_pair(s, ""));
                    else {
                        it = R.find(o);
                        R[o] = make_pair((it->second).first+1, make_pair((it->second).second.first, s));
                    }
                }
            }
            else    return 0;
        }
        map<string, pair<int, string> > Ans;
        map<string, pair<int, string> >::iterator jt;
        for(it = R.begin(); it != R.end(); it++) {
            if((it->second.second).second != "")
                Ans[it->first] = make_pair((it->second).first, (it->second).second.second);
            else
                Ans[it->first] = make_pair((it->second).first, (it->second).second.first);
            //cout << (it->second).second.first << " " << (it->second).second.second << endl;
        }
        printf("Set #%d:\n", ++cases);
        for(jt = Ans.begin(); jt != Ans.end(); jt++) {
            if((jt->second).second != "")
                cout << (jt->second).second;
            else
                cout << jt->first;
            cout << " " << (jt->second).first << endl;
        }
        puts("");
    }
    return 0;
}
/*
xxab xxbb xxba xxaa
#
cd cde
#

Output:
Set #1:
xxaa 2
xxbb 2

Set #2:
cd 1
cde 1
//Wrong the problem description
//we should sort by xx-b and xx-a,
//and output the last ignored word.
*/