题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6208

Here you have a set of strings. A dominator is a string of the set dominating all strings else. The string  SS is dominated by  TT if  SS is a substring of  TT.
InputThe input contains several test cases and the first line provides the total number of cases. 
For each test case, the first line contains an integer  NN indicating the size of the set. 
Each of the following  NN lines describes a string of the set in lowercase. 
The total length of strings in each case has the limit of  100000100000. 
The limit is 30MB for the input file.OutputFor each test case, output a dominator if exist, or No if not.Sample Input 
3
10
you
better
worse
richer
poorer
sickness
health
death
faithfulness
youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness
5
abc
cde
abcde
abcde
bcde
3
aaaaa
aaaab
aaaac
Sample Output 
youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness
abcde
No


题目大意：t次，n个字符串，从中找到一个母串，母串里面包含着其他的字串，输出母串，否则输出no


用到一个函数，就是发现函数，在c++中的string：

#include<stdio.h>
#include<string.h>
#include<math.h>

//#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;

#define ll long long
#define da    0x3f3f3f3f
#define xiao -0x3f3f3f3f
#define clean(a,b) memset(a,b,sizeof(a))// 雷打不动的头文件



string chuan[1000000];

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n,max=xiao,maxi;
		cin>>n;
		int i,j;
		for(i=0;i<n;++i)
			chuan[i].clear();
		for(i=0;i<n;++i)
			cin>>chuan[i];
		string str;
		str=chuan[0];
		for(i=0;i<n;++i)
		{
			if(chuan[i].length()>str.length())
				str=chuan[i];					//找到最长的一个母串 
		}
		int f=1;
		for(i=0;i<n;++i)
		{
			if(str.find(chuan[i])==str.npos)
			{
				f=0;							//只要有一个不符合就 
				break;
			}
		}
		if(f)
			cout<<str<<endl;					//输出对应的 母串 
		else
			cout<<"No"<<endl;
	}
}