题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=51171
题意：给你n个串的集合D，然后求一个长度最短的串s，使得使得s大于等于D中一半的串，又同时小于另一半串。
思路：直接暴力。先对n个串排序，然后选择第n/2-1个串，从第一个字母开始找，在后面添加26个字母枚举，直至找到目标串。
code：
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=1000000;
const int M=2005;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define fr(i,s,n) for (int i=s;i<=n;i++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);string s[M];
int n;
void sol()
{sort(s,s+n);//fr(i,0,n-1) cout<<s[i]<<endl;string s1=s[n/2-1],s2=s[n/2],t="",p="";for (int i=0;;++i){for (char c='A';c<='Z';++c){t=p+c;if(t>=s1&&t<s2){cout<<t<<endl;return;}}p+=s1[i];}
}int main()
{while (~scanf("%d",&n)&&n){fr(i,0,n-1) cin>>s[i];sol();}
}
)
)
