A1077 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

题意：

给定N个 (2≤N≤100)字符串(长度0~256 )，求他们的公共后缀。如果不存在公共后缀，则输出“nai”.

注意：

公共后缀需要从后往前枚举字符串，因此可先将字符串反转，这样问题转换为求N个字符串的公共前缀
求得所有字符串的最短长度 minLen, 枚举在 minLen之内的字符，相同位置的字符相同，则累计
字符串中间存在空格，因此不可用 scanf （因为 scanf 函数的 %s 以空白格包括空格在内进行截断）
getchar() 函数的使用：getchar()是在输入缓冲区顺序读入一个字符(包括空格、回车和Tab) ，前面的scanf()在读取输入时会在缓冲区中留下一个字符'\n'，所以如果不在此加一个getchar()把这个回车符取走的话，getline(）就不会等待从键盘键入字符，而是会直接取走这个“无用的”回车符，从而导致读取有误
题中输入很多字符串，采用C++的string类处理。getline属于string流，接收一个字符串，遇到‘\n’结束。begin()函数返回一个迭代器,指向字符串的第一个元素。end()函数返回一个迭代器，指向字符串的末尾(最后一个字符的下一个位置)。reverse()会将区间[beg,end)内的元素全部逆序（第二个参数是数组最后一个元素的下一个地址）。

#include <iostream>   //需要使用C++中的string类
#include <cstdio>
#include <cstring>
#include <algorithm>  //使用reverse函数
using namespace std;

int n, minLen = 256, ans = 0;
string s[256];    //定义string类型的数组
int main() {
    scanf("%d", &n);
    getchar();
    for(int i = 0; i < n; i++) {
        getline(cin, s[i]);
        int len = s[i].size();
        if(len < minLen) {
            minLen = len;
        }
        reverse(s[i].begin(),s[i].end());
    }
    for(int i = 0; i < minLen; i++) {
        char c = s[0][i];
        bool same = true;
        for(int j = 0; j < n; j++) {
            if(c != s[j][i]) {
                same = false;
                break;
            }
        }
        if(same) {
            ans++;
        } else {
            break;
        }
    }
    if(ans) {
        for(int i = ans - 1; i >= 0; i--) {
            printf("%c", s[0][i]);
        }
    } else {
        printf("nai");
    }
    return 0;
}