PAT甲级——1077.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

题目要求是求相同的字符串后缀

一开始的解题思路是：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
	int N;
	int ans=0;
	scanf("%d",&N);
	int minlen = 256;
	char a[101][300];
	getchar();
	for(int i=0;i<N;i++)
	{
		fgets(a[i],256,stdin);
		int len = strlen(a[i]);
		if(minlen>len) minlen =len;
		for(int j=0;j<minlen/2;j++)//reverse
		{
			char temp;
			temp = a[i][j];
			a[i][j]=a[i][-j+len-1];
			a[i][-j+len-1]=temp;
		}
	}
	for(int i=1;i<minlen;i++)
	{
		bool flag=true;
		char c = a[0][i];
		for(int j = 1;j<N;j++)
		{
			if(c!=a[j][i])
			{
				flag = false;
				break;
			}
		}
		if(flag) ans++;
		else break;
	}
	if(ans)
	{
		for(int i=ans-1;i>=1;i--)
		{
			printf("%c",a[0][i]);
		}

	}
	else
	{
		printf("nai");
	}

	return 0;

}

感觉有点混乱

#include <iostream>
#include <algorithm>
using namespace std;

int main() {
 	int n;
 	scanf("%d\n", &n);
 	string ans;
 	for(int i = 0; i < n; i++) {
 		string s;
 		getline(cin, s);
		 int lens = s.length();
 		reverse(s.begin(), s.end());
		 if(i == 0) {
 				ans = s;
		 		continue;
		 }
        else {
 				int lenans = ans.length();
				int minlen = min(lens, lenans);
 				for(int j = 0; j < minlen; j++) {
 						if(ans[j] != s[j]) {
 						ans = ans.substr(0, j);
 						break;
 						}
		 			}
		 		}
 }
 reverse(ans.begin(), ans.end());
 if (ans.length() == 0)
     ans = "nai";
cout << ans;
return 0;
}

以上的解法更加简便

使用到的新鲜函数是：getline, reverse, min

1. getline

istream& getline (char* s, streamsize n );
istream& getline (char* s, streamsize n, char delim );

字符串的输入方式之一，特点是遇到空格不停止输入

2. reverse

反转范围中元素的顺序[first,last)

3. min

输出两数中较小的一个，同理max是较大那个

巴特西