1077 Kuchiguse （20 分）

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原文链接：http://www.cnblogs.com/Mered1th/p/10428474.html

本文解析了一道名为1077 Kuchiguse的编程题目，介绍了如何通过逆置字符串并寻找最长公共后缀来找出动漫角色的口头禅。示例代码展示了使用C++实现这一算法的过程。

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

Output Specification:

For each test case, print in one line the kuchiguse of the character, i.e., the longest common suffix of all

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 }

转载于:https://www.cnblogs.com/Mered1th/p/10428474.html