PAT 甲级 1077 Kuchiguse

https://pintia.cn/problem-sets/994805342720868352/problems/994805390896644096

 

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

代码:
#include <bits/stdc++.h>
using namespace std;

char s[111][300], out[1111];
int len[111];

int main() {
    int N;
    scanf("%d", &N);
    getchar();
    for(int i = 1; i <= N; i ++) {
        cin.getline(s[i], 300);
    }

    int minn = 333;
    for(int i = 1; i <= N; i ++) {
        len[i] = strlen(s[i]);

        for(int j = 0; j <= len[i] / 2 - 1; j ++)
            swap(s[i][len[i] - j - 1], s[i][j]);

        if(len[i] < minn) {
            minn = len[i];
            continue;
        }
    }

    int cnt = -1;
    for(int j = 0; j < minn; j ++) {
        int flag = 1;
        for(int i = 1; i <= N; i ++) {
            if(s[i][j] != s[1][j]) flag = 0;
        }
        if(flag == 0) break;
        cnt = j;
    }

    if(cnt != -1) {
        for(int i = cnt;i >= 0; i --) {
            printf("%c", s[1][i]);
        }
        printf("\n");
    }
    else
        printf("nai\n");

    return 0;
}

  

posted @ 2018-08-06 22:32  丧心病狂工科女  阅读(457)  评论(0编辑  收藏  举报