PAT 1077 Kuchiguse [一般]

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 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

题目大意：给出m个句子，判断所有句子的最长后缀。

//有空格怎么办？那就不是一个character了。。。

#include <iostream>
#include <algorithm>
#include<cstdio>
#include <vector>
#include<cstring>
#include<string.h>
#include<string>
using namespace std;

int main()
{
    int n;
    cin>>n;
    string pe="",now;//那就以第一个作为标准，剩下的去和它作比较。
    int  j=,k=,mins=j;
    for(int i=;i<n;i++){//如何读取一行，真的是不会。。。。
        getline(cin,now);
        if(pe==""){
            pe=now;continue;
        }
        for(j=pe.size()-,k=now.size()-;k>=,j>=;j--,k--){
            if(pe[j]!=now[k]){
                break;
            }
        }
        if(j<mins){
            mins=j;
        }
    }
    if(mins==){
        cout<<"nai";
    }else{
        for(int i=mins;i<pe.size();i++){
            cout<<pe[i];
        }
    }

    return ;
}

//本来是写成这个样子，根本就不行啊，不过整行读取字符串的函数写的是对的。

根据柳神的修改了之后的，可以AC了

#include <iostream>
#include <algorithm>
#include<cstdio>
#include <vector>

using namespace std;

int main()
{
    int n;
    cin>>n;
    string pe="",now,ans;//那就以第一个作为标准，剩下的去和它作比较。
    getline(cin,now);
    for(int i=;i<n;i++){//如何读取一行，真的是不会。。。。
        getline(cin,now);
        reverse(now.begin(),now.end());
        if(i==){
            ans=now;
        }else{
            int j=;
            for(j=;j<ans.size()&&j<now.size();j++){
                if(ans[j]!=now[j])break;
            }
            ans=ans.substr(,j);
        }
    }
    reverse(ans.begin(),ans.end());
    if(ans=="")
        cout<<"nai";
    else
        cout<<ans;
    return ;
}

1.使用reverse进行倒转，因为是比较后缀，不太好比较，最后的答案需要再翻转过来。

2.关于这个getline(cin,now)读取一行的问题，在输入了n之后，其实就有一个\n被getline给读取进来了，所以这样就浪费了一个，需要单独加一个getline来消灭\n才对！

3.使用substr进行截取答案。