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  • pat1077. 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:
    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
    

    提交代码

    题目其实也是一知半解,但就是AC了。。

     1 #include<iostream>
     2 #include<string>
     3 #include<cstdio>
     4 #include<cstring>
     5 #include<algorithm>
     6 using namespace std;
     7 char line[105][300];
     8 bool cmp(char *a,char *b){
     9     return strlen(a)<strlen(b);
    10 }
    11 int main(){
    12   //freopen("D:\INPUT.txt","r",stdin);
    13   int n,i;
    14   scanf("%d",&n);
    15   getchar();
    16   for(i=0;i<n;i++){
    17     cin.getline(line[i],300);
    18   }
    19   /*for(i=0;i<n;i++){
    20     cout<<line[i]<<endl;
    21   }*/
    22   //sort(line,line+n,cmp);
    23   int j=1,k,len;
    24   char c;
    25   string s="";
    26   for(i=1;i<=strlen(line[0]);i++){
    27     c=line[0][strlen(line[0])-i];
    28     for(k=1;k<n;k++){
    29         len=strlen(line[k]);
    30         if(len<i||line[k][len-i]!=c){
    31             break;
    32         }
    33     }
    34     if(k==n){
    35         s=c+s;
    36     }
    37     else{
    38         break;
    39     }
    40   }
    41   if(s!=""){
    42     cout<<s<<endl;
    43   }
    44   else{
    45     cout<<"nai
    ";
    46   }
    47 }
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  • 原文地址:https://www.cnblogs.com/Deribs4/p/4793605.html
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