二分贪心-B

  • 原题

    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.

  • 思路解析
    题意:
    给你翻译和原词,让你输入原词能输出相应的翻译,如果没有对应的翻译,输出“eh”

  • AC代码

#include<bits/stdc++.h>
using namespace std;
char chip[100001][15];
char bhip[100001][15];
int  idex[100001];
char w[15];
int L,yes;
int cmp(const void*a,const void*b)
{
    return strcmp(bhip[*(int*)a],bhip[*(int*)b]);
}
void main()
{
    int i,j,k,t;
    L=0;
    while(gets(w)&&w[0]!='\0')
    {
        sscanf(w,"%s %s",chip[L],bhip[L]);
        idex[L]=L;
        L++;
    }
    qsort(idex,L,sizeof(idex[0]),cmp);

    while(gets(w))
    {
        i=0;
        j=L-1;
        yes=0;
        while(i<=j)
        {
            k=(i+j)/2;
            t=strcmp(bhip[idex[k]],w);
            if(t>0)
            {
                j=k-1;
            }
            else if(t<0)
            {
                i=k+1;
            }
            else
            {
                yes=1;
                break;
            }

        }
        if(yes==1)
            printf("%s\n",chip[idex[k]]);
        else
        {
            printf("eh\n");
        }

    }
}
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