pat 1077

1077. Kuchiguse (20)

时间限制

100 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

HOU, Qiming

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

提交代码

 

感觉自己弱爆了。必须得强化一下了。

 

#include<iostream>
#include<string>

using namespace std;

string A[101];
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
            getchar();  
            for(int i=0;i<n;i++){
                   getline(cin,A[i]); 
                    }
            
             int flag=1;   
             int cnt=0;    
             for(int i=0;i<256;i++){
                       int len;
                       int len2;
                     for(int j=0;j<n-1;j++){
                             len=A[j].size()-i-1;
                             len2=A[j+1].size()-i-1;
                             if(len<0||len2<0)break;
                             if(A[j][len]!=A[j+1][len2]){
                                    flag=0;
                                    break;
                              }
                              
                             
                             
                             }
                            if(len<0||len2<0)break;
                            if(flag==0)break;
                            else{
                                 cnt++;
                                 }
                     
                     }
                     
                     if(cnt==0){
                        printf("nai\n");}
                     else{
                          int x=A[0].size();
                          
                          for(int i=x-cnt;i<x;i++){
                                  cout<<A[0][i];
                                  }
                          cout<<endl;
      
                          }
         
            }
    

    
    return 0;
    }