PAT A1077. Kuchiguse (20/17)

本文介绍了一种算法,该算法能够从一组日语台词中找出角色特有的语气词(Kuchiguse)。通过对比每句话的结尾部分来寻找共同的模式,如果存在相同的语气词,则输出最长的共同语气词。

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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 <algorithm>
    #include <cmath>
    #include <cstring>
    #define Max 300
    using namespace std;
    struct user
    {
    	char name[11];
    	char pas[11];
    }a[Max],b[Max];
    char getc(char a[],int m)
    {
    	return a[strlen(a)-1-m];
    }
    int main()
    {
    	char la[Max][Max];
    	int n,k=0,f=1;
    	scanf("%d",&n);
    	getchar();
    	for(int i=0;i<n;i++){
    		
    		gets(la[i]);
    	}
    	while (f==1){
    	for(int i=0;i<n-1;i++)
    	{
    		if(getc(la[i+1],k)!=getc(la[i],k)){
    			f=0;
    			break;
    		}
    	}
    	  if(f==1)  k++;
    	  else break;
    	}
    	if(k==0) printf("nai\n");
    	else {
    		for(int i=strlen(la[0])-k;i<strlen(la[0]);i++)
    			printf("%c",la[0][i]);
    	}
    	system("pause");
    	return 0;
    }
    

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