1077. Kuchiguse (20)-PAT甲级真题

1077.
Kuchiguse (20)

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

题目大意：给定N给字符串，求他们的公共后缀，如果不存在公共后缀，就输出“nai”

分析：因为是后缀，反过来比较太麻烦，所以每输入一个字符串，就把它逆序过来再比较，就比较容易了。

首先ans
= s；后来每输入的一个字符串，都和ans比较，如果后面不相同的就把它截取掉。

最后输出ans即可（要逆序输出~~）

#include <iostream>
#include <cstdio>
using namespace std;
int main() {
int n;
scanf("%d", &n);
string ans;
getchar();
for(int i = 0; i < n; i++) {
string s;
getline(cin, s);
int lens = s.length();
for(int j = 0; j < lens / 2; j++) {
swap(s[j], s[lens-1-j]);
}
if(i == 0) {
ans = s;
continue;
} else {
int lenans = ans.length();
int minlen = lens < lenans ? lens : lenans;
for(int j = 0; j < minlen; j++) {
if(ans[j] != s[j]) {
ans = ans.substr(0, j);
break;
}
}
}
}
if(ans.length() == 0) {
printf("nai");
} else {
for(int i = ans.length() - 1; i >= 0; i--) {
printf("%c", ans[i]);
}
}
return 0;
}