1077 Kuchiguse (20 分)

1. 题目

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

2. 题意

找出n个语句中的习惯语。习惯语即每句话结尾都要加上的相同语句。

例：

1111111hello

22hello

33333333333hello

那么这三句话的习惯语为hello。即题目要求找出n句话结尾最长的习惯语，如果没有则输出nai。

3. 思路——字符串

从每句话的末尾遍历，首先记录下第一句话倒数第k个字母，然后与其他几句话的倒数第k个字母进行比较，如果都一致，则将该字母加入结果字符串（即习惯语）。直到出现n句话的倒数第k个字母不一致的情况，查找完毕！其中注意一个细节条件，倒数第k个字母可能对于某句话来说越界了，这时候也要终止查找。

4. 代码

#include <iostream>
#include <string>

using namespace std;

int main()
{
	int n;
	cin >> n;
	string strs[n];
	getchar();
	for (int i = 0; i < n; ++i)
		getline(cin, strs[i]);
		
	int k = 1;
	string res = ""; 
	int flag = 1;		// 终止循环条件 
	char ch;
	while (true)
	{ 
		for (int i = 0; i < n; ++i)
		{
			// 判断每个语句的倒数第k个字母是否相同 
			int j = strs[i].length() - k;
			if (j < 0)	// 越界情况，flag置为0，查找完毕 
			{
				flag = 0;
				break;
			} 
			// 如果i等于0，表示第一个语句，将该字符记录下来，与其他语句对比 
			else if (i == 0)
				ch = strs[i][j];
			// 如果对比结果不一致，说明习惯语收集完毕，退出循环 
			else if (strs[i][j] != ch) 
			{
				flag = 0;
				break;
			}
		}
		if (!flag) break;
		// 因为ch为倒数第k个字符，所以使用ch+res，这样可以按原字符串顺序输出 
		res = ch + res;
		k++;
	}
	if (res != "")
		cout << res << endl;
	else
		cout << "nai" << endl;
	return 0;
}