1077. Kuchiguse (20)

本文介绍了一种通过分析角色对话来识别其特有的句尾表达方式(Kuchiguse)的方法。这种方法通过对多条对话记录进行比对,找出共同的最长句尾部分,以此判断角色的个性特征。

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
    找最大相同后缀,遍历就好

    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<queue>
    #include<vector>
    using namespace std;
    
    int main(){
    	int n;
    	cin>>n;
    	getchar();
    	char s1[1000];
    	gets(s1);
    	int sum=strlen(s1);
    	for(int i=1;i<n;i++){
    		char s2[1000];
    		gets(s2);
    		int cnt=0;
    		for(int j=strlen(s1)-1,k=strlen(s2)-1;j>=0&&k>=0;j--,k--){
    			if(s1[j]==s2[k]){
    				cnt++;
    			}
    			else break;
    		}
    		//strcpy()
    		if(cnt<sum) sum=cnt;
    	}
    	if(sum==0){
    		cout<<"nai";
    	}
    	else{
    		for(int i=strlen(s1)-sum;i<strlen(s1);i++) cout<<s1[i];
    	}
    	return 0;
    }



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