PAT-A 1077.Kuchiguse

寻找最长共同后缀，由于位置固定，难度较小；

A 1077

PAT(Advanced) 1077. Kuchiguse

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

题目描述了一个关于日语表达习惯的背景，主要的要求是：输入N（2<=N<100）个有可能包含空格的字符串（长度小于等于256），找出它们的共同后缀并输出，如果没有共同后缀，则输出nai。

思路

1. 使用c风格字符数组读入字符串

2. 编写一个字符串反转函数，使所有输入的字符串反转，实现后缀的对齐

   //字符串反转函数
   
   void Reverse(char s[], int n) {
       for (int i = 0, j = n - 1; i < j; i++, j--) {
       char temp = s[i];
       s[i] = s[j];
       s[j] = temp;
       }
   }
   

3. 假设输入的第一个字符串是共同后缀subStr

4. 将接下来输入的字符串与subStr逐个字符比较，如果第i个字符不相同，则令subStr[i] = '\0'，相当于截短了子串
5. 如果subStr[0]为'\0'，说明没有共同后缀，输出nai，否则输出subStr

AC代码

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <string.h>

using namespace std;

void Reverse(char s[], int n) {
    for (int i = 0, j = n - 1; i < j; i++, j--) {
        char temp = s[i];
        s[i] = s[j];
        s[j] = temp;
    }
}

int main() {
    int N;
    cin >> N;
    cin.get(); // 抛掉cin流中的'\n'
    char subStr[257];
    for (int i = 0; i < N; i++) {
        char temp[257];
        cin.getline(temp, 257); // 因为输入字符中可能有空格 所以要用getline
        Reverse(temp, strlen(temp));
        if (i == 0) {
            strcpy(subStr, temp);
        }
        else {
            for (int j = 0; j < 257 && subStr[j] != '\0'; j++) {
                if (temp[j] != subStr[j]) {
                    subStr[j] = '\0';
                    break;
                }
            }
        }
    }
    Reverse(subStr,strlen(subStr));
    if (subStr[0] == '\0') {
        cout << "nai";
    }
    else {
        cout << subStr;
    }
    return 0;
}