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  • HDU 1247

    Hat’s Words

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3779    Accepted Submission(s): 1432


    Problem Description
    A hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.
    You are to find all the hat’s words in a dictionary.
     
    Input
    Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 50,000 words.
    Only one case.
     
    Output
    Your output should contain all the hat’s words, one per line, in alphabetical order.
     
    Sample Input
    a ahat hat hatword hziee word
     
    Sample Output
    ahat hatword
     1 //大致题意: 判断一个单词是否由其他两个单词组成 
     2 #include <iostream>
     3 #include <vector>
     4 #include <string>
     5 #include <cstring>
     6 #include <map>
     7 using namespace std;
     8 int main()
     9 {
    10     int i,j,k;
    11     string s="";
    12     vector <string > v;
    13     map <string ,int > mm;
    14     while(cin>>s)
    15     {
    16         v.push_back(s);
    17         mm[s]++;//不为0表示存在 
    18         s.clear();
    19     }
    20     for(i=0;i<v.size();i++)
    21     {
    22         string s1="";
    23         s1=v[i];
    24         string s2 = "";//不可放在内层 
    25         for(j=0;j+1<s1.size();j++)
    26         {
    27             s2 += s1[j];
    28             string s3 = s1.substr(j+1);
    29             if(mm[s2]&&mm[s3])
    30             {
    31                 cout<<s1<<endl;
    32                 s2.clear();
    33                 break;//可能一个单词有多种匹配,满足一种输出即可 
    34             }
    35         }
    36     }
    37     system("pause");
    38     return 0;
    39 }
    40                 
    41             
    42             
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  • 原文地址:https://www.cnblogs.com/hxsyl/p/2647319.html
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