PAT(A) 1077. Kuchiguse (20)

最新推荐文章于 2020-03-12 22:53:22 发布

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最新推荐文章于 2020-03-12 22:53:22 发布

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原文链接：http://www.cnblogs.com/claremore/p/6548959.html

本文介绍了一种通过分析角色对话来识别其口语习惯的方法。利用最长公共后缀算法，可以找出动漫或漫画中角色特有的口语结尾，即“Kuchiguse”。通过实际样例展示了如何从多个句子中提取共同的口语特征。

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

#include <cstdio>
#include <cstring>
/*求最长公共后缀
(1)取n个字符串中的最短长度minlen
(2)将接收到的n个字符串逆序，转化为求公共前缀
(3)求公共前缀，只需对比字符串前minlen长度的字符（即下标范围：0 -> minlen-1）
*/
int main()
{
    int n;
    char str[105][260];     //至多100个字符串，每个字符串至多256个字符            
    scanf("%d", &n);
    getchar();              //划重点(1)：接受掉换行，不然会被下面的gets()收走

    //(1)反转字符串s[i], 转化为求公共前缀
    int minlen=256;         //划重点(2)：n个字符串中取最短长度，初始化为最大
    for(int i=0; i<n; i++){
        //注：用scanf输入会错，因为无法判断字符串有效长度
        gets(str[i]);    　　//gets()会吃掉空格,换行符，故前面须用getchar()/getc()接收

        int len=strlen(str[i]);
        if(len<minlen)
            minlen=len;       //取最小长度
        for(int j=0; j<len/2; j++){      //反转字符串str[i], 转化为求公共前缀
            char temp=str[i][j];
            str[i][j]=str[i][len-1-j];
            str[i][len-1-j]=temp;
        }
    }
    //(2)判断所有字符串的第i个字符是否相等
    int ans=0;    //记录公共前缀有多少个字符相等
    for(int i=0; i<minlen; i++){
        char c=str[0][i];           //取第一个字符串的第i个字符
        bool same=true;
        for(int j=1; j<n; j++){  　　//判断其余字符串的第i个字符是否等于c
            if(str[j][i]!=c){       //只要有一个不等于c，说明公共前缀到此为止
                same=false;
                break;
            }
        }
        if(same)    ans++;     //若所有字符串的第i位相等时，公共前缀长度ans+1
        else    break;
    }
    //(3)倒序输出公共前缀 => 公共后缀
    if(ans){
        for(int i=ans-1; i>=0; i--)
            printf("%c", str[0][i]);
    }
    else
        printf("nai");          //不存在公共前缀

    return 0;
}