Codeforces Round #170 (Div. 2) B. New Problem（好题）

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本文探讨了一种算法，用于找出最短且未出现在过往题目中的字符串，作为新题目的名称。通过遍历所有可能的一到两个字符组合，并检查它们是否已存在于输入的题目集合中，从而找到符合条件的最短原创题名。

Coming up with a new problem isn't as easy as many people think. Sometimes it is hard enough to name it. We'll consider a title original if it doesn't occur as a substring in any titles of recent Codeforces problems.

You've got the titles of n last problems — the strings, consisting of lowercase English letters. Your task is to find the shortest original title for the new problem. If there are multiple such titles, choose the lexicographically minimum one. Note, that title of the problem can't be an empty string.

A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s₁s₂... s_|s| (where |s| is the length of string s) is string s_ls_l + 1... s_r.

String x = x₁x₂... x_p is lexicographically smaller than string y = y₁y₂... y_q, if either p < q and x₁ = y₁, x₂ = y₂, ... , x_p = y_p, or there exists such number r (r < p, r < q), that x₁ = y₁, x₂ = y₂, ... , x_r = y_r and x_r + 1 < y_r + 1. The string characters are compared by their ASCII codes.

Input

The first line contains integer n (1 ≤ n ≤ 30) — the number of titles you've got to consider. Then follow n problem titles, one per line. Each title only consists of lowercase English letters (specifically, it doesn't contain any spaces) and has the length from 1 to 20, inclusive.

Output

Print a string, consisting of lowercase English letters — the lexicographically minimum shortest original title.

Sample test(s)

Input

5
threehorses
goodsubstrings
secret
primematrix
beautifulyear

Output

Input

4
aa
bdefghijklmn
opqrstuvwxyz
c

Output

ab

Note

In the first sample the first 9 letters of the English alphabet (a, b, c, d, e, f, g, h, i) occur in the problem titles, so the answer is letter j.

In the second sample the titles contain 26 English letters, so the shortest original title cannot have length 1. Title aa occurs as a substring in the first title.

想了各种数据结构去做，看了大神的思路才知道自己弱爆了……明明暴力就能轻松解决……

The total number of different strings of 2 letters is 26² = 676, but the total length of the input strings is no more than 600. It means that the length of answer is no more than 2. So just check all the strings of length 1 and 2.

 1 #include <iostream>
 2 #include <string>
 3 #include <set>
 4 #include <map>
 5 #include <vector>
 6 #include <stack>
 7 #include <queue>
 8 #include <cmath>
 9 #include <cstdio>
10 #include <cstring>
11 #include <algorithm>
12 using namespace std;
13 #define LL long long
14 #define cti const int
15 #define ctll const long long
16 #define dg(i) cout << "*" << i << endl;
17 
18 bool ha[2000];
19 char s[31][21];
20 int Hash(char a, char b)
21 {
22     return (a - 'a') * 50 + b - 'a' + 26;
23 }
24 
25 int main()
26 {
27     int n;
28     while(scanf("%d", &n) != EOF)
29     {
30         memset(ha, false, sizeof(ha));
31         while(n--)
32         {
33             scanf("%s", s[n]);
34             for(int i = 0; s[n][i] != '\0'; i++)
35             {
36                 ha[s[n][i]-'a'] = true;
37                 if(s[n][i+1] != '\0') ha[Hash(s[n][i], s[n][i+1])] = true;
38             }
39         }
40         int ok = 0;
41         for(int i = 0; i < 26; i++)
42             if(!ha[i])
43             {
44                 printf("%c\n", i + 'a');
45                 ok = 1;
46                 break;
47             }
48         if(!ok)
49         {
50             char x, y;
51             for(x = 'a'; x <= 'z' && !ok; x++)
52             {
53                 for(y = 'a'; y <= 'z' && !ok; y++)
54                 {
55                     if(!ha[Hash(x, y)])
56                         ok = 1;
57                 }
58             }
59             printf("%c%c\n", x - 1, y - 1);
60         }
61     }
62     return 0;
63 }