After improving the marketing strategy, TIANKENG has made a fortune and he is going to step into the status of TuHao. Nevertheless, TIANKENG wants his restaurant to go international, so he decides to name his restaurant in English. For the lack of English skills, TIANKENG turns to CC, an English expert, to help him think of a property name. CC is a algorithm lover other than English, so he gives a long string S to TIANKENG. The string S only contains eight kinds of letters——-‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’, ‘H’. TIANKENG wants his restaurant’s name to be out of ordinary, so the restaurant’s name is a string T which should satisfy the following conditions: The string T should be as short as possible, if there are more than one strings which have the same shortest length, you should choose the string which has the minimum lexicographic order. Could you help TIANKENG get the name as soon as possible?

Meanwhile, T is different from all the substrings of S. Could you help TIANKENG get the name as soon as possible?
Input
The first line input file contains an integer T(T<=50) indicating the number of case.
In each test case:
Input a string S. the length of S is not large than 1000000.
Output
For each test case:
Output the string t satisfying the condition.(T also only contains eight kinds of letters——-‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’, ‘H’.)
Sample Input
3
ABCDEFGH
AAABAACADAEAFAGAH
ACAC
Sample Output
AA
BB
B

題意：
給定一個由A~H8種字元組成，長度不超過1000000的字串。問在這個字串的最短不存在連續子串。若存在多個最短連續子串，輸出字典序最小的一個。
題解：
先估計最長的答案是多少？我的估計是6個字元長，為什麼?長度為6的字串共有8^6≈250000，如果不存在公用的情況，需要長度250000*6的長度，已經長過1000000了。（有可能是7個字元長的，但7個字元的樣例必須考慮字元共用，很難弄出來，所以6個字元就夠了，如果想要保證正確性，可以用7個字元的）。
然後找出所有長度小於等於6的所有子串，由於子串是連續的，所以只用6*n的時間去查詢。用f[i][j]標記存在的子串，i表示長度，j表示字元的雜湊值（字元種類只有8個，用i位8進位制就可以表示所有的情況）。最後遍歷f陣列，找出第一個未被標記的子串即可。

#include <bits/stdc++.h>
using namespace std;
int vis[9000005];
char s[1000005];
char zimu[9]={'A','B','C','D','E','F','G','H'};
int len;
char ans[1000005];
int solve(int x)
{
    memset(vis,0,sizeof(vis));
    for(int i=0;i<=len-x;i++)
    {
        int k=0;
        int sum=0;
        while(k<x)
        {
            sum=sum*8+s[i+k]-'A';
            k++;
        }
        vis[sum]=1;
    }
    int t=pow(8,x);
    int kk=0;
    for(int i=0;i<t;i++)
    {
        if(!vis[i]) 
        {
            int kk=x;
            while(x)
            {
                int tt=i%8;
                ans[x]=zimu[tt];
                i/=8;
                x--;
            }
            for(int j=1;j<=kk;j++)
                printf("%c",ans[j] );
            return 1;
        }
    }
    return 0;
}


int main()
{
    int t;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        gets(s);
        len=strlen(s);
        for(int i=1;!solve(i);i++) ;
            printf("\n");
    }
}