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.
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.
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
/*
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;
}