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。
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 ...”。
GaoXZh
Magi
Einst
Quark
LaoLao
FatMouse
ZhaShen
fantacy
latesum
SenSen
QuanQuan
whatever
whenever
Potaty
hahaha
.
Magi and Potaty are inviting you to dinner...
LaoLao
FatMouse
whoever
.
FatMouse is the only one for you...
LaoLao
.
Momo... No one is for you ...
&:也是服了自己了,少了个点没看出来,Orz
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
char s[555555][50];
int main()
{
int i = 1;
while(1)
{
scanf("%s",s[i]);
getchar();
if(s[i][0] == '.') break;
i ++;
}
if(i <= 2){
printf("Momo... No one is for you ...\n");
}
else if(i < 14){
printf("%s is the only one for you...\n",s[2]);
}
else if(i >= 14){
printf("%s and %s are inviting you to dinner...\n",s[2],s[14]);
}
return 0;
}