PAT甲级.1077. Kuchiguse (20)

1077. Kuchiguse (20)

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题目

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?

输入格式

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.

输出格式

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”.

输入样例1

3
Itai nyan~
Ninjin wa iyadanyan~
uhhh nyan~

输出样例1

nyan~

输入样例2

3
Itai!
Ninjinnwaiyada T_T
T_T

输出样例2

nai

PAT链接

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思路

1.取第一个读取的字符串反转后存入ans[]内，之后对每个读入的字符串反转之后与ans从头到后比较，发现不同字符跳出，将之前相同的字符串再更新到ans中
2.将ans反转输出

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代码

/*** @tag     PAT_A_1077* @authors R11happy ([e-mail])* @date    2016-8-30 23:56-00:30* @version 1.0* @Language C++* @Ranking  380/2402* @function null*/#include <cstdio>#include <cstdlib>#include <cstring>char str[110][260];//反转字符串void reverse(char s[]){    int len = strlen(s);    for (int i = 0; i<len / 2; i++)    {        int tmp = s[i];        s[i] = s[len - 1 - i];        s[len - 1 - i] = tmp;    }}int main(int argc, char const *argv[]){    int N,j;    char ans[260];    scanf("%d", &N);    getchar();    gets(ans);    reverse(ans);    for (int i = 1; i<N; i++)    {        char tmp[260];        gets(str[i]);        reverse(str[i]);        int len_ans = strlen(ans);        int len_str = strlen(str[i]);        int len = len_ans > len_str ? len_str : len_ans;        for (j = 0; j<len; j++)        {            if (ans[j] == str[i][j])    tmp[j] = ans[j];            else break;        }        tmp[j] = '\0';  //注意，不是tmp[++j]        strcpy(ans, tmp);    }    reverse(ans);    if (strlen(ans)) puts(ans);    else    printf("nai\n");    return 0;}

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收获

1.对于从后向前比较的字符串，先反转再操作的话会比较简单

//反转字符串void reverse(char s[]){    int len = strlen(s);    for (int i = 0; i<len / 2; i++)    {        int tmp = s[i];        s[i] = s[len - 1 - i];        s[len - 1 - i] = tmp;    }}

2.字符串注意处理最后的’\0’

        tmp[j] = '\0';  //注意，不是tmp[++j]