1077 Kuchiguse （20 分）

Posted 2021-02-10 mered1th

1077 Kuchiguse （20 分）

The Japanese language is notorious for its sentence ending particles. Personal preference of such particles can be considered as a reflection of the speaker‘s personality. Such a preference is called "Kuchiguse" and is often exaggerated artistically in Anime and Manga. For example, the artificial sentence ending particle "nyan~" is often used as a stereotype for characters with a cat-like personality:

• Itai nyan~ (It hurts, nyan~)

• Ninjin wa iyada nyan~ (I hate carrots, nyan~)

Now given a few lines spoken by the same character, can you find her Kuchiguse?

Input Specification:

Each input file contains one test case. For each case, the first line is an integer N (2≤N≤100). Following are N file lines of 0~256 (inclusive) characters in length, each representing a character‘s spoken line. The spoken lines are case sensitive.

Output Specification:

For each test case, print in one line the kuchiguse of the character, i.e., the longest common suffix of all N lines. If there is no such suffix, write nai.

Sample Input 1:

3
Itai nyan~
Ninjin wa iyadanyan~
uhhh nyan~

Sample Output 1:

nyan~

Sample Input 2:

3
Itai!
Ninjinnwaiyada T_T
T_T

Sample Output 2:

nai


分析： 求字符串的公共后缀。思路是先逆置字符串，然后从第一个开始比较。

 1 /**
 2 * Copyright(c)
 3 * All rights reserved.
 4 * Author : Mered1th
 5 * Date : 2019-02-24-21.51.09
 6 * Description : A1077
 7 */
 8 #include<cstdio>
 9 #include<cstring>
10 #include<iostream>
11 #include<cmath>
12 #include<algorithm>
13 #include<string>
14 #include<unordered_set>
15 #include<map>
16 #include<vector>
17 #include<set>
18 using namespace std;
19 const int maxn=110;
20 string a[maxn];
21 int main(){
22 #ifdef ONLINE_JUDGE
23 #else
24     freopen("1.txt", "r", stdin);
25 #endif
26     int n,minlen=10000;
27     cin>>n;
28     getchar();
29     for(int i=0;i<n;i++){
30         getline(cin,a[i]);
31         int len=a[i].size();
32         if(minlen>len) minlen=len;
33         reverse(a[i].begin(),a[i].end());
34     }
35     string ans="";
36     for(int i=0;i<minlen;i++){
37         char c=a[0][i];
38         bool same=true;
39         for(int j=1;j<n;j++){
40             if(c!=a[j][i]){
41                 same=false;
42                 break;
43             }
44         }
45         if(same) ans+=c;
46         else break;
47     }
48     if(ans.size()){
49         reverse(ans.begin(),ans.end());
50         cout<<ans;
51     }
52     else{
53         cout<<"nai";
54     }
55     return 0;
56 }