pat-1077 Kuchiguse(20)（字符串中寻找最长公共后缀）

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

作者: HOU, Qiming

单位: 浙江大学

时间限制: 150 ms

内存限制: 64 MB

代码长度限制: 16 KB

 重点：过滤输入n后的换行符！

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <set>
using namespace std;

int main()
{


    int n;
    scanf("%d",&n);
    getchar();//重要！！！否则就不对！
    string res;//res得到公共最长后缀
    for(int i = 0 ; i < n;i++)
    {
        string str;
        getline(cin,str);
        
        int len = str.length();
        reverse(str.begin(),str.end());//翻转字符串，从最后一个开始比较
        if(i == 0)
        {
            res = str;//先把公共最长后缀默认为第一条输入的字符串
            continue;
        }
        else
        {
            int lenres = res.length();
            int minlen = min(lenres,len);
            for(int j = 0;j < minlen;j++)
            {
                if(res[j] != str[j])
                {
                    res = res.substr(0,j);//到第j个不相等了，最长公共部分就是之前匹配的，截取下来
                    break;
                }
            }
        }
    }
    reverse(res.begin(),res.end());//翻转输出
    if(res.length() == 0)
        res = "nai";
    cout<<res;

    return 0;
}