Hat’s Words(map,string)

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

Sponsor

 题意：给出好多串，判断某个串能否由给出的串中的某两个串拼接而成。

思路：两个for循环。

利用map的find函数，以及string的截取功能。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<map>
using namespace std;
map<string,int>kk;
map<string,int>::iterator it;
int main()
{
	string s;
	while(cin>>s)
	{
		kk[s]=1;
	}
	for(it=kk.begin();it!=kk.end();it++)//遍历每个串 
	{
		string a=it->first;// a为里面的参数 
		for(int j=1;j<a.size()-1;j++)//枚举分割点
		{
			string a1(a,0,j);//从0开始往后j位 
			string a2(a,j);
			if(kk.find(a1)!=kk.end()&&kk.find(a2)!=kk.end())
			{
				cout<<a<<endl;
				break;
			}
		}
	}
	return 0;
}