codeforces 278B New Problem 字串尋找

B. New Problemtime limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 titleoriginal 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 = s_{1s2... s|s| (where |s| is
the length of string s) is string slsl + 1... sr.}

String x = x_{1x2... xp is lexicographically
smaller than string y = y1y2... yq,
if either p < q and x1 = y1, x2 = y2, ...
, xp = yp,
or there exists such number r (r < p, r < q),
that x1 = y1, x2 = y2, ...
, xr = yr and xr + 1 < yr + 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

5threehorsesgoodsubstringssecretprimematrixbeautifulyear

output

j

input

4aabdefghijklmnopqrstuvwxyzc

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.

感受： 這道題當時看到了覺得好複雜~~~  看瞭解題報告後發現自己進入了思維的誤區~~

題意： 從給出的n個字串中找到第一次沒有出現的字串（先從字串的長度1開始遍曆a-z），然後從字串的長度2開始，遍曆（aa-zz）然後就結束了~~

為什麼遍曆到zz就結束了呢，因為最多30個字串每個字串的長度為20，然後最長也就600個字元，然後連續的兩兩一組超不過aa-zz的全部組合，然後由aa-zz 總共為26*26=676個了 也就是出先不了3個字元的字串~~~汗~~就是兩個以內一定能找到。

然後正確的選用庫函數也是其中的關鍵的一環，尋找用strchr和strstr來尋找吧，有簡單又有效率~~

post code：

#include<stdio.h>#include<string.h>char a[40][30];char ch;char ans[4];int n;int find1(){   for(ch='a';ch<='z';ch++){     int judge=0;             for(int i=1;i<=n;i++)       if(strchr(a[i],ch)!=NULL) judge=1; //找到字元      if(judge==0)return 1;      // 沒要找到字元 輸出答案      }   return 0;}int find2(){    ans[0]='a';    while(ans[0]<='z'){        for(ans[1]='a';ans[1]<='z';ans[1]++){            int judge=0;            for(int i=1;i<=n;i++)              if(strstr(a[i],ans)!=NULL) judge=1;           if(judge==0)return 1;        }         ans[0]++;     }    return 0;}int main(){    while(scanf("%d",&n)!=EOF){         int flag=0;         int i;         for(i=1;i<=n;i++)            scanf("%s",a[i]);         flag=find1(); //尋找字串的長度為1的         if(flag==1){            printf("%c\n",ch);continue;          }          flag=find2(); //尋找字串的長度為2的         if(flag==1){            printf("%s\n",ans);continue;             }             }   return 0;}