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  • 1077. Kuchiguse (20)

    题目如下:

    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
    



题目要求从一系列给定字符串中找出公共结尾,因为这个结尾将会被所有的字符串所共有,所以我们保存第一个输入的字符串,用它来和各个字符串比较,找出共同位置,一旦发现没有共同部分,立即返回nai,因为输入和输出流是分开的,所以我们提前返回得到的输出文件没有任何问题;如果发现了公共部分,就要进行保存,我们用一个suffix变量保存这个公共字串,当发现的公共部分长度比suffix小,说明公共部分缩短,应该更新,为了保证第一次的公共部分能够写入suffix,我们把suffix的值初始化为第一次输入。

对于有空格的输入,应该使用getline(),getline()有个问题是会吃掉上一行的回车作为一个空行,因此getline()之前如果有输入一定要getchar()吃掉那个回车再处理

代码如下:

#include <iostream>
#include <stdio.h>
#include <string>
#include <vector>

using namespace std;

int main()
{
    int N;
    string temp,suffix,input;
    cin >> N;
    getchar();
    getline(cin,temp);
    suffix = temp;
    for(int i = 1; i < N; i++){
        getline(cin,input);
        int tempCur = temp.length() - 1;
        int inputCur = input.length() - 1;
        while(input[inputCur] == temp[tempCur]){
            inputCur--;
            tempCur--;
        }
        int len = temp.length() - tempCur - 1;
        if(len == 0){
            suffix = "nai";
            break;
        }
        if(suffix.length() > len){
            suffix = temp.substr(tempCur + 1);
        }
    }
    cout << suffix << endl;
    return 0;
}


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  • 原文地址:https://www.cnblogs.com/aiwz/p/6154046.html
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