L1-035 情人节

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,我决定造福你们。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。ios

输入格式:

输入按照点赞的前后顺序给出不知道多少个点赞的人名,每一个人名占一行,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。spa

输出格式:

根据点赞状况在一行中输出结论:若存在第2我的A和第14我的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 ...”。code

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<iostream> #include<string>
using namespace std; int main() { int cnt = 0; string str, A, B; while (cin >> str) { if (str == ".") break; else { cnt++; if (cnt == 2) A = str; if (cnt == 14) B = str; } } if (cnt >= 14) cout << A << " and " << B << " are inviting you to dinner..."; if (cnt >= 2 && cnt < 14) cout << A << " is the only one for you..."; if (cnt < 2) cout << "Momo... No one is for you ..."; return 0; }
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