PAT (Advanced Level) Practise 1077 Kuchiguse (20)

最新推荐文章于 2021-11-20 10:03:05 发布

Wang_128

最新推荐文章于 2021-11-20 10:03:05 发布

阅读量209

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT/PTA甲级 ----水题

本文链接：https://blog.youkuaiyun.com/a664607530/article/details/74529111

----水题同时被 2 个专栏收录

234 篇文章

订阅专栏

PAT/PTA甲级

174 篇文章

订阅专栏

本文介绍了一道关于日本语言特点“Kuchiguse”的识别问题。任务要求从多个句子中找出共同的句尾特征，并提供了完整的代码实现，用于识别特定角色的语言习惯。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1077. Kuchiguse (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HOU, Qiming

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：给你n个字符串，判断是不是有标准结尾，有就输出，没有就输出nai

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n, len[200];
char ch[200][300];
string ans;

int main()
{
	while (~scanf("%d", &n))
	{
		getchar();
		ans = "";
		for (int i = 0; i < n; i++) gets(ch[i]), len[i] = strlen(ch[i]);
		for (int i = 1; i; i++)
		{
			int flag = 1;
			for (int j = 1; j < n; j++)
			{
				if (len[j] < i) { flag = 0; break;}
				if (ch[j][len[j] - i] != ch[j - 1][len[j - 1] - i]) {flag = 0; break;}
			}
			if (flag) ans = ch[0][len[0]-i]+ans;
			else break;
		}
		if (ans == "") ans = "nai";
		cout << ans << endl;
	}
	return 0;
}