A1077 Kuchiguse

2020-08-10 11:04:21 3 分钟读完 (大约 507 个字)

Categories:

technology

Tags: BeInvisible, C/C++, PAT, 数据结构, 算法

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

Code:

#pragma warning(disable: 4996)
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

using namespace std;

int main(int argc, char *argv[]) {
	int N;
	scanf("%d", &N);
	char **niHonJin = new char*[N];//日本人和日语都是NiHonJin
	int *len = new int[N];
	getchar();//很怪，因为没法用gets，用gets_s读的时候会多读一个字符
	for (int i = 0; i < N; i++) {
		niHonJin[i] = new char[256];
		//gets_s(niHonJin[i], 300);//PAT的编译器不支持这个，我吐了
		//scanf("%[^\n]", niHonJin[i]);
		int j = 0;
		while (((niHonJin[i][j] = getchar()) != '\n'))//来了，最原始的读数据写法
			j++;
		if (j == 0) {//出现空字符串，但其实测试点没有这种情况，最后一个测试点到底是啥？
			printf("nai");
			system("pause");
			return 0;
		}
		niHonJin[i][j] = '\0';
		len[i] = strlen(niHonJin[i]);
	}
	char suffix[256];
	char buff;
	int suff = -1;
	int maxLen = *max_element(len, len + N);//algorithm
	for (int i = 1; i <= 256; i++) {
		buff = niHonJin[0][len[0] - i];
		for (int j = 0; j < N; j++) {
			if (buff == niHonJin[j][len[j] - i])
				continue;
			else
				suff = i - 1;
		}
		if (suff != -1)//这个判断条件要在上面，因为可能maxLen内的最后一位(也就是str[0])是不一样的
			break;
		if (i == maxLen) {//最后一个测试点是这样测出来的2 123456 123465
			suff = maxLen;
			break;
		}
	}
	//printf("%d", suff);//right
	strcpy(suffix, niHonJin[0] + len[0] - suff);
	if (suff == 0)
		printf("nai");
	else
		printf("%s", suffix);
	system("pause");
	return 0;
}
//书上思路是反转字符串，我也写一个试试看

A1077 Kuchiguse

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