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  • POJ 2503, Babelfish

    Time Limit: 3000MS  Memory Limit: 65536K
    Total Submissions: 12536  Accepted: 5529


    Description
    You have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have a dictionary to help you understand them.

     

    Input
    Input consists of up to 100,000 dictionary entries, followed by a blank line, followed by a message of up to 100,000 words. Each dictionary entry is a line containing an English word, followed by a space and a foreign language word. No foreign word appears more than once in the dictionary. The message is a sequence of words in the foreign language, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.

    Output
    Output is the message translated to English, one word per line. Foreign words not in the dictionary should be translated as "eh".

    Sample Input
    dog ogday
    cat atcay
    pig igpay
    froot ootfray
    loops oopslay

    atcay
    ittenkay
    oopslay

    Sample Output
    cat
    eh
    loops

    Hint
    Huge input and output,scanf and printf are recommended.

    Source
    Waterloo local 2001.09.22


    // POJ2503.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    using namespace std;

    struct Word
    {
        Word():next(
    0){}
        
    char from[12], to[12];
        Word
    * next;
    };
    inline 
    int calcKey(char word[],int SIZE)
    {
        
    int key = 0;
        
    int K = strlen(word);
        
    for(int j=0; j<K; ++j)
            key
    =((key<<2)+(((int)word[j])>>4))^(((int)word[j])<<10);
        key 
    = key % SIZE;
        key 
    = key < 0 ? key + SIZE : key;
        
    return key;
    }

    int main(int argc, char* argv[])
    {
        
    const int SIZE = 100003;
        Word hash[SIZE];
        
    char w[25];

        
    while(gets(w)&& w[0!= 0)
        {
            Word
    * pw = new Word;
            sscanf(w,
    "%s %s",pw->to,pw->from);
            
    int key = calcKey(pw->from, SIZE);
            Word
    * pt = &hash[key];
            
    while (pt->next != NULL) pt = pt->next;
            pt
    ->next = pw;
        }

        
    while(gets(w)&& w[0!= 0)
        {
            
    int key = calcKey(w, SIZE);
            Word
    * pt = &hash[key];
            
    bool found = false;
            
    while (pt->next != NULL)
            {
                pt 
    = pt->next;
                
    if (strcmp(pt->from, w)==0)
                {
                    found 
    = true;
                    
    break;
                }
            }
            
    if(found) printf("%s\n",pt->to);
            
    else printf("eh\n");
        }

        
    return 0;
    }

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  • 原文地址:https://www.cnblogs.com/asuran/p/1579598.html
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