HDOJ 1247 Hat’s Words

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8325    Accepted Submission(s): 3003

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

aahathathatwordhzieeword

 

Sample Output

ahathatword

/*HDOJ 1247 查找单词是否是其他2个组成 */#include<iostream>#include<stdio.h>#include<map>#include<string>using namespace std;char str[50001][100];map<string,int> ss;int main(){    int k,k1,k2,i,j;    char f[100],e[100];    k=0;    //    freopen("test.txt","r",stdin);        while(scanf("%s",str[k])!=EOF)    {        ss[str[k]]=5;//表示单词的存在         k++;        }        for(i=0;i<k;i++)    {        for(j=0;str[i][j]!='\0';j++)//将单词拆成2部分，                                         {                            //分别查看是否存在与map中                        f[j]=str[i][j];f[j+1]='\0';            if(ss[f]==5)//前一个单词存在             {                for(k1=j+1,k2=0;str[i][k1]!='\0';k1++,k2++)                    e[k2]=str[i][k1];                e[k2]='\0';                if(ss[e]==5)                {                    printf("%s\n",str[i]);                    break;                }            }        }    }    return 0;}