[转]PAT甲级练习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

求最后的相同字符串，直接看了别人的，这里转载一下，地址http://www.liuchuo.net/archives/2065

题目大意：给定N给字符串，求他们的公共后缀，如果不存在公共后缀，就输出“nai”
分析：因为是后缀，反过来比较太麻烦，所以每输入一个字符串，就把它逆序过来再比较，就比较容易了。
首先ans = s；后来每输入的一个字符串，都和ans比较，如果后面不相同的就把它截取掉。
最后输出ans即可（要逆序输出~~）

#include <iostream>#include <cstdio>using namespace std;int main() {    int n;    scanf("%d", &n);    string ans;    getchar();    for(int i = 0; i < n; i++) {        string s;        getline(cin, s);        int lens = s.length();        for(int j = 0; j < lens / 2; j++) {            swap(s[j], s[lens-1-j]);        }        if(i == 0) {            ans = s;            continue;        } else {            int lenans = ans.length();            int minlen = lens < lenans ? lens : lenans;            for(int j = 0; j < minlen; j++) {                if(ans[j] != s[j]) {                    ans = ans.substr(0, j);                    break;                }            }        }    }    if(ans.length() == 0) {        printf("nai");    } else {        for(int i = ans.length() - 1; i >= 0; i--) {            printf("%c", ans[i]);        }    }    return 0;}