团体程序设计天梯赛-练习集-L1-035. 情人节

L1-035. 情人节

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 ...

 

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 100010;
 5 int main() {
 6     string s;
 7     int i = 0;
 8     string a, b;
 9     int flag0 = 0, flag1 = 0;
10     while (cin >> s) {
11         if (s == ".")
12             break;
13         i += 1;
14         if (i == 2) {
15             flag0 = 1;
16             a = s;
17         }
18         if (i == 14) {
19             b = s;
20             flag1 = 1;
21         }
22     }
23     if (flag0 == 0 && flag1 == 0) {
24         printf ("Momo... No one is for you ...\n");
25     } else if (flag0 == 1 && flag1 == 0) {
26         cout << a;
27         printf (" is the only one for you...\n");
28     } else {
29         cout << a << " and " << b;
30         printf (" are inviting you to dinner...\n");
31     }
32     return 0;
33 }

 

posted @ 2018-03-20 20:20  April_AA  阅读(456)  评论(0编辑  收藏  举报