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 ...
题目思路不难,下面分别给出我两次写题的代码
#include<bits/stdc++.h>
using namespace std;
int main()
{
int i=0;
string s[999];
for(;;i++)
{
cin>>s[i];
if(s[i]==".")
{
i--;//去掉读"."加的次数
break;
}
}
if(i<1)
{
cout<<"Momo... No one is for you ...";
}
else if(i<13)
{
cout<<s[1]<<" is the only one for you...";
}
else
{
cout<<s[1]<<" and "<<s[13]<<" are inviting you to dinner...";
}
return 0;
}
#include<bits/stdc++.h>
using namespace std;
int main()
{
int cnt=0;
string a,b;
while(1)
{
string s;
cin>>s;
if(s==".") break;
cnt++;
if(cnt==2)
{
a=s;
}
if(cnt==14)
{
b=s;
}
}
if(cnt<2) cout<<"Momo... No one is for you ...";
else if(cnt<14) cout<<a<<" is the only one for you...";
else cout<<a<<" and "<<b<<" are inviting you to dinner...";
return 0;
}