PAT1077: Kuchiguse

最新推荐文章于 2024-07-29 04:27:15 发布

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最新推荐文章于 2024-07-29 04:27:15 发布

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原文链接：http://www.cnblogs.com/0kk470/p/7775021.html

本文介绍了一种用于识别动漫角色口头禅(Kuchiguse)的算法，通过寻找多个字符串的最长公共后缀来确定角色的独特语言习惯。该算法适用于不超过100条的输入数据，并使用C++实现。

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1077. Kuchiguse (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HOU, Qiming

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

思路

求输入的几个字符串的最长公共后缀。
用一个cmstr暂时记录最长公共后缀，然后对于每一个输入的新字符串循环比较更新就行。

代码

#include<iostream>
using namespace std;
int main()
{
    int N;
    cin >> N;
    string cmstr;
    getchar();
    for(int i = 0; i < N; i++)
    {
        string s;
        getline(cin,s);
        const int len = s.size();
        for(int j = 0; j < len/2; j++)
        {
            swap(s[j],s[len - j - 1]);
        }
        if(i == 0)
        {
            cmstr = s;
            continue;
        }
        else
        {
            int minlen = len > cmstr.size()?cmstr.size():len;
            for(int v = 0; v < minlen; v++)
            {
                if(cmstr[v] != s[v])
                {
                    cmstr = cmstr.substr(0,v);
                    break;
                }
            }
        }

    }
    if(cmstr.size() == 0)
        cout << "nai" <<endl;
    else
    {
        for(int i = cmstr.size() - 1;i >= 0;i--)
        {
            cout << cmstr[i];
        }
    }
}

转载于:https://www.cnblogs.com/0kk470/p/7775021.html