A1077 Kuchiguse Suffix-后缀

Posted 2021-03-10 pennyxia

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). 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

思路：

• 输入n个字符串，存到字符串数组中，将所有字符串进行reverse,然后从第一个元素开始查找，直到n个字符串对应位置不一样时停止；

• 将最后结果截取存到str中，翻转后输出。

问题：

问题1：最后一行无法输入

问题2：无法正常输出“nai”

解决：

解决1：getline（）之前需要用getchar（）吸收cin>>n之后的回车

解决2：if (i == 1)cout << "nai";改成if (i == 0)cout << "nai";

因为for循环的运行是这样的：

我们把其命名为表达式1、2、3。for执行时首先执行表达式1,然后执行表达式2,如果循环成立,在循环结束后,下一个循环前执行表达式3,然后再执行表达式2进行判断。

所以，在该题中条件不成立时在循环中break，即跳出循环，所以不会再执行i++；如果是正常结束循环，会执行i++，然后判断表达式2是否成立判断是否需要结束循环。

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 using namespace std;
 5 int main() {
 6     int n,minlen=256;
 7     string s[105];
 8     cin >> n;
 9     getchar();
10     //scanf_s("%d", &n);
11     for (int i = 0; i < n; i++) {
12         getline(cin, s[i]);
13         reverse(s[i].begin(), s[i].end());
14         if (s[i].length() < minlen)
15             minlen = s[i].length();
16     }
17     int i=0,tag = 0;
18     for (i = 0; i < minlen; i++) {
19         for (int j = 1; j < n; j++) {
20             if (s[j][i] != s[0][i]) {
21                 tag = 1;
22                 break;
23             }
24         }
25         if (tag == 1)break;
26     }
27     if (i == 0)cout << "nai";
28     else {
29         string str= s[0].substr(0, i);
30         reverse(str.begin(), str.end());
31         cout << str;
32     }
33     return 0;
34 }