A1077. Kuchiguse

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

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
void rev(char a[]){
    int len = strlen(a);
    for(int i = 0; i < len / 2; i++)
        swap(a[i],a[len - 1 - i]);
}
int main(){
    int N, minL = 99999, tag = 1, len = 0;
    char str[100][1000], sen[1000];
    scanf("%d", &N);
    getchar();
    for(int i = 0; i < N; i++){
        gets(str[i]);
        rev(str[i]);
        if(minL > strlen(str[i]))
            minL = strlen(str[i]);
    }
    for(int i = 0; i < minL; i++){
        char c = str[0][i];
        for(int j = 0; j < N; j++)
            if(str[j][i] != c){
                tag = 0;
                break;
            }
        if(tag == 0)
            break;
        len++;
    }
    if(len != 0)
        for(int j = len - 1; j >= 0; j--)
            printf("%c", str[0][j]);
    else
        printf("nai");
    cin >> N;
    return 0;
}

总结：

1、本题即求公共后缀，由于每个字符串长度不一，可以在输入字符串后逆序，转换为求公共前缀。

2、关于字符串的处理，本题无法使用scanf的%s读入字符串，因为字符串中存在空格。所以可以用gets()，在遇到回车时结束。但要注意，在scanf读入N后，还存在一个回车键，需要被吸收掉，否则会导致gets()读到一个回车空串。

3、由于PAT不支持gets()读入带空格的一行。以后遇到复杂的字符串题时，使用string而不是char[]。