PAT 1077. Kuchiguse (20)

1077. Kuchiguse (20)

时间限制

100 ms

内存限制

65536 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:

3Itai nyan~Ninjin wa iyadanyan~uhhh nyan~

Sample Output 1:

nyan~

Sample Input 2:

3Itai!Ninjinnwaiyada T_TT_T

Sample Output 2:

nai

题意就是找所有句子的最长公共结尾。坑点在于公共结尾允许是以空格开头的

#include <iostream>#include<stdio.h>#include<queue>#include<string.h>#include<string>#include<vector>using namespace std;string sentence[105];int main(){    int n;    cin>>n;    getchar();    for(int i=0;i<n;i++)      getline(cin,sentence[i]);    int flag=1;    int cnt=1;    char path[10000];    int p=0;    while(flag)    {        int l=sentence[n-1].size()-cnt;        char now=sentence[n-1][l];        for(int i=n-2;i>=0;i--)        {            int x=sentence[i].size()-cnt;            if(x<0) {                       flag=0;break;                    }            if(sentence[i][x]!=now)            {               flag=0;               break;            }        }          if(!flag) break;          path[p++]=now;          cnt++;    }      //while(path[p-1]==' '&&p>0) p--;   去公共开头的空格      if(p==0) cout<<"nai";      else      {          p--;          for(;p>=0;p--)            cout<<path[p];      }   return 0;}