1077. Kuchiguse (20)字符串处理——PAT (Advanced Level) Practise

Posted 2020-10-02

题目信息

1077. Kuchiguse (20)

时间限制100 ms
内存限制65536 kB
代码长度限制16000 B

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

解题思路

反转字符串后逐个字符比較

AC代码

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main()
{
    int n;
    vector<string> v;
    string s, rs;
    cin >> n;
    getline(cin, s);// 去掉回车
    while (n--){
        getline(cin, s);
        reverse(s.begin(), s.end());
        v.push_back(s);
    }
    int loc = -1;
    bool flag = true;
    while (flag && ++loc < v[0].size()){
        for (int i = 0; i < v.size(); ++i){
            if (v[i][loc] != v[0][loc]){
                flag = false;
                break;
            }
        }
        if (flag){
            rs += v[0][loc];
        }
    }
    if (rs.empty()){
        cout <<"nai" <<endl;
    }else{
        reverse(rs.begin(), rs.end());
        cout <<rs <<endl;
    }
    return 0;
}