1.题目要求求每个句子后面的公共字符串，如果有则输出，如果没有则输出nai。
2.需要使用getline输入，避免空格断开。
3.其他没有太多特点的地方。
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
AC代码：
[c language=”++”]
//#include<string>
//#include <iomanip>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;
int main(void)
{
string nStr;
getline(cin, nStr);
int n = 0;
for (int i = 0; i < nStr.size(); i++)
n = n * 10 + nStr[i] – ‘0’;
vector<string> sentence(n);
int minLen = INT_MAX;
for (int i = 0; i < n; i++)
{
getline(cin,sentence[i]);
minLen = min(minLen,(int) sentence[i].size());
}
string ans = "";
bool diff = false;
for (int i = 1; i <= minLen; i++)
{
char c = sentence[0][sentence[0].size() – i];
for (int j = 1; j < sentence.size(); j++)
{
if (c != sentence[j][sentence[j].size() – i])
{
diff = true;
break;
}
}
if (diff) break;
else ans = c + ans;
}
if (ans.size() == 0)
cout << "nai" << endl;
else
cout << ans << endl;
return 0;
}
[/c]