1077 Kuchiguse （20 分)

2019-04-07

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

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

题目大意

给出一串字符串，求所有字符串的最大后缀

分析

按照题意，从尾到头判断各字符串上相应位置字符是否相同即可。后缀初始化为空，如果有两字符不同且后缀为空，那么输出nai，否则输出相应后缀。当最短的字符串已经遍历完，都没有发现不同，也输出后缀。这题通过率0.2，很奇怪，感觉应该不止。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;

string s[105];
int main() {
    int n,minl=9999;
    cin>>n;
    getchar();
    for(int i=0;i<n;i++) {
        getline(cin, s[i]);
        if(s[i].length()<minl)
            minl=s[i].length();
    }
    string temp="";
    for(int i=1;i<=minl;i++){
        char c=s[0][s[0].length()-i];
        for(int j=1;j<n;j++){
            if(c!=s[j][s[j].length()-i]){
                if(temp=="")
                    cout<<"nai";
                else
                    cout<<temp;
                return 0;
            }
        }
        temp=c+temp;
    }
    cout<<temp;
    return 0;
}