zoukankan      html  css  js  c++  java
  • L1-035 情人节 (15 分)

    L1-035 情人节 (15 分)

    以上是朋友圈中一奇葩贴:“2月14情人节了,我决定造福大家。第2个赞和第14个赞的,我介绍你俩认识…………咱三吃饭…你俩请…”。现给出此贴下点赞的朋友名单,请你找出那两位要请客的倒霉蛋。

    输入格式:

    输入按照点赞的先后顺序给出不知道多少个点赞的人名,每个人名占一行,为不超过10个英文字母的非空单词,以回车结束。一个英文句点.标志输入的结束,这个符号不算在点赞名单里。

    输出格式:

    根据点赞情况在一行中输出结论:若存在第2个人A和第14个人B,则输出“A and B are inviting you to dinner...”;若只有A没有B,则输出“A is the only one for you...”;若连A都没有,则输出“Momo... No one is for you ...”。

    输入样例1:

    GaoXZh
    Magi
    Einst
    Quark
    LaoLao
    FatMouse
    ZhaShen
    fantacy
    latesum
    SenSen
    QuanQuan
    whatever
    whenever
    Potaty
    hahaha
    .
    

    输出样例1:

    Magi and Potaty are inviting you to dinner...
    

    输入样例2:

    LaoLao
    FatMouse
    whoever
    .
    

    输出样例2:

    FatMouse is the only one for you...
    

    输入样例3:

    LaoLao
    .
    

    输出样例3:

    Momo... No one is for you ...


    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
    //建立string a[15]

    //i++循环
    //若i>14,则置为14
    //输入a[i]
    //若输入a[i]为 . ,break

    //判断i大小,三情况
    string a[15];
    int i = -1;
    while(1)
    {
    i++;
    if(i > 14) i = 14;
    cin>>a[i];
    if(a[i][0] == '.') break;
    }
    if(i < 2)
    cout<<"Momo... No one is for you ...";
    else if(i == 14)
    cout<<a[1]<<" and "<<a[13]<<" are inviting you to dinner...";
    else
    cout<<a[1]<<" is the only one for you...";
    return 0;
    }

     
  • 相关阅读:
    ZOJ Problem Set–2417 Lowest Bit
    ZOJ Problem Set–1402 Magnificent Meatballs
    ZOJ Problem Set–1292 Integer Inquiry
    ZOJ Problem Set–1109 Language of FatMouse
    ZOJ Problem Set–1295 Reverse Text
    ZOJ Problem Set–1712 Skew Binary
    ZOJ Problem Set–1151 Word Reversal
    ZOJ Problem Set–1494 Climbing Worm
    ZOJ Problem Set–1251 Box of Bricks
    ZOJ Problem Set–1205 Martian Addition
  • 原文地址:https://www.cnblogs.com/hear0/p/10539367.html
Copyright © 2011-2022 走看看