A1077 Kuchiguse Suffix-后缀

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。

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main() {
    int n,minlen=256;
    string s[105];
    cin >> n;
    getchar();
    //scanf_s("%d", &n);
    for (int i = 0; i < n; i++) {
        getline(cin, s[i]);
        reverse(s[i].begin(), s[i].end());
        if (s[i].length() < minlen)
            minlen = s[i].length();
    }
    int i=0,tag = 0;
    for (i = 0; i < minlen; i++) {
        for (int j = 1; j < n; j++) {
            if (s[j][i] != s[0][i]) {
                tag = 1;
                break;
            }
        }
        if (tag == 1)break;
    }
    if (i == 0)cout << "nai";
    else {
        string str= s[0].substr(0, i);
        reverse(str.begin(), str.end());
        cout << str;
    }
    return 0;
}