PAT (Advanced Level) Practise 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~)
• 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_TSample Output 2: nai
  找一下最长公共后缀，直接暴力判断就好了。

#include<cstdio>
#include<stack>
#include<cstring>
#include<string>
#include<vector>
#include<queue>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
const int INF = 0x7FFFFFFF;
const int maxn = 1e3 + 15;
int n, len[maxn];
char s[maxn][maxn];
string ans;

int main()
{
  scanf("%d", &n);  getchar();
  for (int i = 0; i < n; i++) gets(s[i]), len[i] = strlen(s[i]);
  for (int i = 1; i; i++)
  {
    int flag = 1;
    for (int j = 1; j < n; j++)
    {
      if (len[j] < i) { flag = 0; break; }
      if (s[j][len[j] - i] != s[j - 1][len[j - 1] - i])
      {
        flag = 0; break;
      }
    }
    if (flag) ans = s[0][len[0] - i] + ans; else break;
  }
  if (ans == "") ans = "nai";
  cout << ans << endl;
  return 0;
}