1077. Kuchiguse (20)-PAT甲级真题

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会比较容易～
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～，所以先将ans倒置reverse一下～）

#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();
            if(lens < lenans) swap(ans, s);
            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;
}

#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();

if(lens < lenans) swap(ans, s);

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;

}

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