PAT A1077

最新推荐文章于 2021-09-17 16:35:03 发布

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最新推荐文章于 2021-09-17 16:35:03 发布

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本文介绍了一个算法，用于从一组动漫或漫画中角色的台词中找出最长的共同结尾部分，即所谓的‘口头禅’。该算法首先读取多行输入，然后通过反转字符串并逐字符比较来确定所有台词共有的最长后缀。

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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
```

实现代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char s[105][266];
void reverse(char * a){
    char ch;
    for(int i = 0, j = strlen(a) - 1; i < j; i++, j--){
        ch = a[i];
        a[i] = a[j];
        a[j] = ch;
    }
}
int main(){
    int N, min_len = 266;
    cin>>N;
    getchar();
    for(int i = 0; i < N; i++){
        gets(s[i]);
        min_len > strlen(s[i]) ? min_len = strlen(s[i]) : min_len = min_len;
        reverse(s[i]);
    }
    char ch; bool k; int jl = 0;
    for(int i = 0; i < min_len; i++){
        k = true;
        ch = s[0][i];
        for(int j = 1; j < N; j++){
            if(ch != s[j][i]){
                k = false;
                break;
            }
        }
        if(k) jl++;
        else break;
    }
    if(jl){
        for(int i = jl - 1; i >= 0; i--){
            cout<<s[0][i];
        }
    }else{
        cout<<"nai"<<endl;
    }
    return 0;
}