題意：給你一些字串，如果有字串是其他兩個字串的聯合。輸出這個字串

思路：先把字典樹造出來，然後利用substr分割成兩塊暴力查詢即可

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include <sstream>
#include<vector>
#include
 
<cmath>    
#include<stack>
#include<time.h>
#include<ctime>
using namespace std;
#define LD long double
#define LL long long
#define PI acos(-1.0)
#define E 1e-9
#define eps 1e-7
#define INF 0x3f3f3f3f
#define inf 1<<30
#define maxn 100005
string a[50009] = {};
int n = 1, num = 1
 
;
int tire[50000][30] = {};
int vis[500009] = {};
void build(string s)
{
    int len = s.length();
    int p = 0;
    for (int i = 0; i < len; i++)
    {
        int k = s[i] - 'a';
        if (!tire[p][k])
        {
            tire[p][k] = num;
            num++;
        }
        p = tire[p][k];
    }
    vis[p] 
 
= 1;
}
int search(string s)
{
    int len = s.length();
    int p = 0;
    for (int i = 0; i < len; i++)
    {
        int k = s[i] - 'a';
        if (!tire[p][k])
        {
            return 0;
        }
        p = tire[p][k];
    }
    if (vis[p] == 1)
    {
        return 1;
    }
    else
    {
        return 0;
    }
}
int main()
{
    while (cin>>a[n])
    {
        
        build(a[n]);
        n++;
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j < a[i].length(); j++)
        {
            string s1, s2;
            s1 = a[i].substr(0, j);
            s2 = a[i].substr(s1.length(), a[i].length()-s1.length());
            if (search(s1) && search(s2))
            {
                cout << a[i] << endl;
                break;
            }
            else
            {
                continue;
            }
        }
    }
}