zoukankan      html  css  js  c++  java
  • Babelfish(二分)

    Babelfish
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 37238   Accepted: 15879

    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
    题解:此处用到了sscanf的用法,很巧妙;
    代码:
     1 #include<stdio.h>
     2 #include<algorithm>
     3 #include<string.h>
     4 using namespace std;
     5 const int MAXN=100010;
     6 struct Node{
     7     char en[20],ei[20];//无语开小了。。。 
     8 };
     9 Node dt[MAXN];
    10 int cmp(Node a,Node b){
    11     return strcmp(a.ei,b.ei)<0;
    12 }
    13 int erfen(char *x,int l,int r){
    14     while(l<=r){
    15         int mid=(l+r)>>1;
    16         if(!strcmp(dt[mid].ei,x))return mid;
    17         if(strcmp(dt[mid].ei,x)>0)r=mid-1;
    18         else l=mid+1;
    19     }
    20     return -1;
    21 }
    22 int main(){
    23     int dm=0;
    24     char str[50];
    25     Node a;
    26     while(gets(str)){
    27         if(str[0]=='')break;
    28             sscanf(str,"%s%s",a.en,a.ei);//这里很巧妙。。。 
    29             dt[dm++]=a;
    30             }
    31             sort(dt,dt+dm,cmp);
    32             while(~scanf("%s",str)){
    33             int t=erfen(str,0,dm-1);
    34             if(t==-1)puts("eh");
    35             else printf("%s
    ",dt[t].en);
    36             }
    37     return 0;
    38 }
  • 相关阅读:
    HDU 5861 Road (线段树)
    HDU 5857 Median (推导)
    HDU 5858 Hard problem (数学推导)
    HDU 5867 Water problem (模拟)
    UVALive 7455 Linear Ecosystem (高斯消元)
    A bug about RecipientEditTextView
    当Activity出现Exception时是如何处理的?
    FontSize sp 和 dp 的区别
    Android的Overlay机制
    关于控件问题的分析
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4836607.html
Copyright © 2011-2022 走看看