1077 Kuchiguse (20分)

原创于 2020-02-11 23:07:34 发布 · 117 阅读

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本文探讨了日语中独特的句子结尾粒子，这些粒子反映了说话者的个性。通过分析动漫和漫画中角色的语言习惯，即“Kuchiguse”，文章提供了一个算法，用于识别并找出角色对话中的共同后缀，以此来确定其语言偏好。

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

题意：找最大相同的后缀

思路：用vector容器储存字符串，两个循环搞定，外循环，所有字符串最小长度，内循环，每个字符串的最后一位字符是否相等，维护一个string来保存相等的字符，只要有一个不相等，直接跳出两层循环，然后判断string是否为空，若为空，则说明没有相同的后缀，反之，将string逆序输出即可～

#include<iostream>

#include<vector>
#include<algorithm>


using namespace std;

int main(){
    int n;
    cin >> n;
    vector<string> vec(n);
    char c;
    c = getchar();
    int min_len = 0x7fffffff;
    for(int i = 0; i < n; i++){

        getline(cin, vec[i]);
        min_len = min(min_len, int(vec[i].length()));
    }
    string s = "";
    char temp;
    bool  flag = true;
    for(int j = 1; j <= min_len; j++){
        for(int i = 0; i < n - 1; i++){

            if(vec[i][vec[i].length() - j] != vec[i + 1][vec[i + 1].length() - j]){
                flag = false;
                break;
            }

            temp = vec[i][vec[i].length() - j];
        }
        if(flag){
            s += temp;
        }else
        {
            break;
        }
        
            


    }
    
    reverse(s.begin(), s.end());
    if(s == ""){
        cout << "nai" << endl;
    }else
    {
        cout << s << endl;
    }
    
    

    return 0;
}