《ACM程序设计》书中题目B

最新推荐文章于 2022-07-08 00:46:13 发布

TimEckel

最新推荐文章于 2022-07-08 00:46:13 发布

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分类专栏： acm水题文章标签： acm

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本文介绍了一个使用C++实现的简单程序，该程序能够将一种虚构的语言（FatMouse语言）翻译成英语。通过构建字典映射关系，程序能够处理多达100,005个词汇的翻译任务。对于未出现在字典中的词汇，默认翻译为‘eh’。文章提供了完整的源代码，并简要提及了实现过程中的一些注意事项。

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Description
We all know that FatMouse doesn’t speak English. But now he has to be prepared since our nation will join WTO soon. Thanks to Turing we have computers to help him.

Input Specification

Input consists of up to 100,005 dictionary entries, followed by a blank line, followed by a message of up to 100,005 words. Each dictionary entry is a line containing an English word, followed by a space and a FatMouse word. No FatMouse word appears more than once in the dictionary. The message is a sequence of words in the language of FatMouse, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.

Output Specification

Output is the message translated to English, one word per line. FatMouse 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
Output for Sample Input

cat
eh
loops

这题思路还可以，就是具体实现的时候会遇到问题，以为不习惯用map，开始遇到了小麻烦，读输入，判断，输出，别忘了判断空行。
代码如下：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    map<string, int>ma;
    char s1[20], s2[20], s3[20];

    char s[100007][15];
    int e=1;    
    while(scanf("%c", &s1[0]) && s1[0]!='\n' )
    {
        scanf("%s %s%*c", s1+1, s2 );
        ma[s2]=e++ ;
        strcpy(s[e-1], s1);
    }
    int flag;
    while(scanf("%s", s3)!=EOF)
    {
        if(ma[s3]>=1)
        {
            printf("%s\n", s[ma[s3]] );
        }
        if(ma[s3]==0)
        {
            printf("eh\n");
        }
    }    
    return 0;
}