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首先,先贴柳神的博客

https://www.liuchuo.net/ 这是地址

想要刷好PTA,强烈推荐柳神的博客,和算法笔记

题目原题

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

生词如下:

notorious 声名狼藉的

particles 粒子

stereotype 模式,模板

题目大意

就是给你若干个字符串,要你找出这些字符串的相同的尾缀

思路如下

① 把我们输入的字符串逆置先

② 在一个一个找他们相同的尾缀

代码如下

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
int main(void) {string a[101];int N,i=0,min=257,flag=0;cin >> N;int N1 = N;cin.ignore();while (N--) {getline(cin, a[i]);if (min > a[i].length())min = a[i].length();reverse(a[i].begin(), a[i].end());i++;}i = 0;int ans = 0;for(; i < min; i++) {char b = a[0][i];for (int j = 1; j < N1; j++) {if (b != a[j][i]) {flag = 1;break;}}if (flag) {break;}else {ans++;}}if (ans==0)cout << "nai";else for (int j = ans - 1; j >= 0; j--) {cout << a[0][j];}return 0;
}

下面请和我一起欣赏婼神的代码

#include <iostream>
#include <algorithm>
#include<cstring>
#include<string>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;
}