• 1077. Kuchiguse


    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:
    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
     1 #include<stdio.h>
     2 #include<math.h>
     3 #include<stdlib.h>
     4 #include<string.h>
     5 
     6 int main()
     7 {
     8     int n, i, j, k = 0;
     9     char str[110][300];
    10     scanf("%d", &n);
    11     getchar();
    12     int len, minlen = 300;
    13     for(i = 0; i < n; i++)
    14     {
    15         gets(str[i]);
    16         len = strlen(str[i]);
    17         if(len < minlen)
    18         {
    19             minlen = len;
    20         }
    21         for(j = 0; j < len / 2; j++)
    22         {
    23             char temp = str[i][j];
    24             str[i][j] = str[i][len - 1 - j];
    25             str[i][len - 1 - j] = temp;
    26         }
    27     }
    28     for(j = 0; j < minlen; j++)
    29     {
    30         int flag = 1;
    31         char temp = str[0][j];
    32         for(i = 0; i < n; i++)
    33         {
    34             if(str[i][j] != temp)
    35             {
    36                 flag = 0;
    37                 break;
    38             }
    39         }
    40         if(flag)
    41             k++;
    42         else
    43             break;
    44     }
    45     if(k <= 0)
    46         printf("nai
    ");
    47     else
    48     {
    49         for(j = k - 1; j >= 0; j--)
    50         {
    51             printf("%c", str[0][j]);
    52         }
    53         printf("
    ");
    54     }
    55     return 0;
    56 
    57 }
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  • 原文地址:https://www.cnblogs.com/yomman/p/4275937.html
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