0
点赞
收藏
分享

微信扫一扫

L1-035 情人节 (15 分)

流计算Alink 2022-03-12 阅读 48
c++

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()
{
    string s;
    int count = 0, flag = 0;
    while(cin >> s && s != "."){
        count++;
        if(count==2) {
            cout << s << ' ';
            flag = 1;
        }
        if(count==14) {
            cout << "and " << s << ' ' << "are inviting you to dinner...";
            return 0;
        }
    }
    if(flag == 1)
        cout << "is the only one for you...";
    else
        cout << "Momo... No one is for you ...";
    return 0;
}

 

举报

相关推荐

0 条评论