以上是朋友圈中一奇葩貼:“2月14情人節了,我決定造福大家。第2個贊和第14個讚的,我介紹你倆認識…………咱三吃飯…你倆請…”。現給出此貼下點讚的朋友名單,請你找出那兩位要請客的倒黴蛋。
輸入格式:
輸入按照點讚的先後順序給出不知道多少個點讚的人名,每個人名佔一行,爲不超過10個英文字母的非空單詞,以回車結束。一個英文句點.
標誌輸入的結束,這個符號不算在點贊名單裏。
輸出格式:
根據點贊情況在一行中輸出結論:若存在第2個人A和第14個人B,則輸出“A and B are inviting you to dinner...”;若只有A沒有B,則輸出“A is the only one for you...”;若連A都沒有,則輸出“Momo... No one is for you ...”。
輸入樣例1:
GaoXZh
Magi
Einst
Quark
LaoLao
FatMouse
ZhaShen
fantacy
latesum
SenSen
QuanQuan
whatever
whenever
Potaty
hahaha
.
輸出樣例1:
Magi and Potaty are inviting you to dinner...
輸入樣例2:
LaoLao
FatMouse
whoever
.
輸出樣例2:
FatMouse is the only one for you...
輸入樣例3:
LaoLao
.
輸出樣例3:
Momo... No one is for you ...
源碼:
#include<string>
#include<iostream>
using namespace std;
int main(){
string s, a, b;
int count = 1;
cin>>s;
while(s.find(".") == string::npos){
cin>>s;
if(s.find(".") != string::npos) break;
count++;
if(count == 2) a = s;
else if(count == 14) b = s;
}
if(count < 2) cout<<"Momo... No one is for you ...";
else if(count <14) cout<<a<<" is the only one for you...";
else cout<<a<<" and "<<b<<" are inviting you to dinner...";
return 0;
}