PAT(甲级)1077 Kuchiguse (20 分)

最新推荐文章于 2021-11-28 20:30:04 发布

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该博客介绍了日本语言中著名的句子结尾助词，并提出这些助词可以反映说话者的个性，称为'Kuchiguse'。文章以动漫和漫画中的例子说明，并提供了一道编程题，要求找出同一角色所有台词的最长公共后缀。题目给出了输入输出规格和样例，并提醒注意输入处理的细节。

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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

简析：这道题主要的意思，就是输出最长相同的尾缀，如果尾缀最后的部分都不一样，就输出nai。总的思想很简单，不过这道题有个问题，就是输入n后，需要getchar()，不然第一个字符串是个“”。

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    cin >> n;
    getchar();
    vector<string> str(n);
    int minlen = 256;
    for (int i = 0; i < n;i++)
    {
        getline(cin,str[i]);
        if(str[i].length() < minlen)
            minlen = str[i].length();
        reverse(str[i].begin(),str[i].end());
    }
    int l = 0;
    for (int i = 0; i < minlen;i++)
    {
        char c = str[0][i];
        bool flag = 1;
        for (int j = 1; j < n;j++)
        {
            if(c != str[j][i])
            {
                flag = 0;
                break;
            }
        }
        if(flag == 0)
            break;
        l++;
    }
    if(l == 0)
        cout << "nai";
    else
    {
        string str1 = str[0].substr(0, l);
        reverse(str1.begin(), str1.end());
        cout << str1;
    }
    return 0;
}