zoukankan      html  css  js  c++  java
  • PAT Advanced 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


    #include <iostream>
    #include <algorithm>
    #include <stack>
    using namespace std;
    
    int main()
    {
        int T,min=1000;
        cin>>T;
        string s[T];
        getline(cin,s[0]);
        stack<char> sta;
        for(int i=0;i<T;i++){
            getline(cin,s[i]);
            reverse(s[i].begin(),s[i].end());
            if(min>s[i].length()) min=s[i].length();
        }
        bool isSame;
        for(int i=0;i<min;i++){
            isSame=true;
            for(int j=0;j<T;j++){
                if(s[j][i]!=s[0][i]) isSame=false;
            }
            if(isSame) sta.push(s[0][i]);
            else break;
            /**不相同直接break掉,这边是一个失误点
            失误案例   abddnai
                        aaaanai*/
        }
        if(sta.empty()) cout<<"nai";
        while(!sta.empty()){
            cout<<sta.top();
            sta.pop();
        }
        system("pause");
        return 0;
    }
  • 相关阅读:
    求解:块级元素的宽度自适应问题
    list 小练习
    codevs1017乘积最大
    codevs1048石子归并
    luogu1387 最大正方形
    BZOJ1305: [CQOI2009]dance跳舞
    linux下分卷tar.bz文件的合并并解压缩
    ubuntu命令查补
    认识与学习BASH(中)
    认识与学习BASH
  • 原文地址:https://www.cnblogs.com/littlepage/p/11306661.html
Copyright © 2011-2022 走看看