• PAT 甲级 1077 Kuchiguse (20 分)(简单,找最大相同后缀)


    1077 Kuchiguse (20 分)
     

    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). 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

    题意:

    题意是寻找一组字符串中最大相同后缀,并输出。如果没有:则输出‘nai’

    题解:

    简单题,可以reverse反转一下找最大前缀

    AC代码:

    #include<bits/stdc++.h>
    using namespace std;
    string a[105];
    int main(){
        int n;
        cin>>n;
        getchar();
        for(int i=1;i<=n;i++){
            getline(cin,a[i]);
            reverse(a[i].begin(),a[i].end());
            //cout<<a[i]<<endl;
        }
        string s="";
        int f=1; 
        char c;
        for(int i=0;i<280;i++){
            for(int j=1;j<=n;j++){
                if(i>=a[j].length()){
                    f=0;
                    break;
                }
                if(j>=2){
                    if(a[j][i]!=a[1][i]){
                        f=0;
                        break;
                    }
                }else{
                    c=a[1][i];
                }
            }
            if(f==1){
                s+=c;
            }
        }
        reverse(s.begin(),s.end());
        if(s!="") cout<<s;
        else cout<<"nai";
        return 0;
    } 
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  • 原文地址:https://www.cnblogs.com/caiyishuai/p/13270584.html
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